Free Math Worksheets

- Compare the numerator with the derivative of the denominator. - Try a substitution using the denominator. - Recall the antiderivative of \(\frac{1}{u}\). - When using substitution in a definite integral, update the limits or substitute back before evaluating.

1. Let \(u=x^3+x+1\). Then \(du=(3x^2+1)\,dx\), and \((6x^2+2)\,dx=2\,du\). 2. Change the limits: when \(x=0\), \(u=1\); when \(x=1\), \(u=3\). 3. The integral becomes \(2\int_1^3 \frac{1}{u}\,du=2[\ln|u|]_1^3=2\ln(3)-2\ln(1)=2\ln(3)\).

\(2\ln(3)\)

- Compare the factor outside the exponential with the derivative of its exponent. - What substitution uses the exponent of \(e\)? - Remember the negative sign in the derivative of cosine. - Update the limits after making a substitution in a definite integral.

1. Let \(u=\cos(x)\), so \(du=-\sin(x)\,dx\). 2. Change the limits: when \(x=0\), \(u=1\); when \(x=\pi\), \(u=-1\). 3. Then \(\int_0^\pi \sin(x)e^{\cos(x)}\,dx=-\int_1^{-1}e^u\,du=\int_{-1}^{1}e^u\,du\). 4. Evaluate: \([e^u]_{-1}^{1}=e-e^{-1}=e-\frac{1}{e}\).

\(e-\frac{1}{e}\)

- Use the exponent of \(e\) as a possible substitution. - Compare its derivative with the remaining factor in the integrand. - Update the limits after substituting. - A negative sign can be handled by reversing the new limits.

1. Let \(u=\frac{1}{x}\). Then \(du=-\frac{1}{x^2}\,dx\). 2. Change the limits: when \(x=1\), \(u=1\); when \(x=2\), \(u=\frac{1}{2}\). 3. The integral becomes \(-\int_1^{\frac{1}{2}}e^u\,du=\int_{\frac{1}{2}}^1e^u\,du\). 4. Evaluate: \([e^u]_{\frac{1}{2}}^1=e-e^{\frac{1}{2}}=e-\sqrt{e}\).

\(e-\sqrt{e}\)

- In a), compare the numerator with the derivative of the denominator. - Recall the antiderivative associated with \(\frac{u'}{u}\). - In b), rewrite the square root using exponent \(\frac{1}{2}\). - Include the reciprocal of the inner derivative when integrating a linear composite function.

1. For a), let \(u=x^2+x\), so \(du=(2x+1)\,dx\). Then \(\int_1^4\frac{2x+1}{x^2+x}\,dx=[\ln(x^2+x)]_1^4=\ln(20)-\ln(2)=\ln(10)\). 2. For b), let \(u=2x+1\), so \(du=2\,dx\). Change the limits: \(x=0\) gives \(u=1\), and \(x=4\) gives \(u=9\): \(\int_0^4\sqrt{2x+1}\,dx=\frac{1}{2}\int_1^9u^{\frac{1}{2}}\,du\). 3. Evaluate: \(\frac{1}{2}[\frac{2}{3}u^{\frac{3}{2}}]_1^9=\frac{1}{3}(27-1)=\frac{26}{3}\).

a) \(\ln(10)\) b) \(\frac{26}{3}\)

- In a), compare the numerator with the derivative of the denominator. - Recall the logarithmic antiderivative pattern. - In b), identify the derivative of the inner linear expression. - What function differentiates to sine, and how must the inner derivative be handled?

1. For a), the derivative of \(x^2-3x+10\) is \(2x-3\), and \(6x-9=3(2x-3)\). Therefore, \(F(x)=3\ln|x^2-3x+10|\). 2. For b), account for the inner derivative \(\frac{1}{2}\). Since an antiderivative of sine is negative cosine, \(G(t)=-16\cos\left(\frac{t}{2}+\pi\right)\).

a) \(F(x)=3\ln|x^2-3x+10|\) b) \(G(t)=-16\cos\left(\frac{t}{2}+\pi\right)\)

- In a), compare the derivative of the exponent with the factor outside the exponential. - Look for a chain-rule pattern in each integrand. - In b), rewrite the expression as \([\ln(x)]^2\cdot\frac{1}{x}\). - Use the inner function as the substitution variable.

1. For a), let \(u=1-x^2\), so \(du=-2x\,dx\). Then \(\int xe^{1-x^2}\,dx=-\frac{1}{2}\int e^u\,du=-\frac{1}{2}e^{1-x^2}+C\). 2. For b), let \(u=\ln(x)\), so \(du=\frac{1}{x}\,dx\). Then \(\int\frac{[\ln(x)]^2}{x}\,dx=\int u^2\,du=\frac{1}{3}u^3+C=\frac{1}{3}[\ln(x)]^3+C\).

a) \(F(x)=-\frac{1}{2}e^{1-x^2}\) b) \(H(x)=\frac{1}{3}[\ln(x)]^3\)

- Compare the numerator with the derivative of the denominator. - Use the denominator as a substitution. - Recall the antiderivative of \(\frac{1}{u}\). - Use a logarithm property to combine the final difference.

1. Let \(u=x^3-x+5\). Then \(du=(3x^2-1)\,dx\), and \((6x^2-2)\,dx=2\,du\). 2. Change the limits: when \(x=1\), \(u=5\); when \(x=2\), \(u=11\). 3. The integral becomes \(2\int_5^{11}\frac{1}{u}\,du=2[\ln|u|]_5^{11}\). 4. Therefore, the value is \(2\ln(11)-2\ln(5)=2\ln\left(\frac{11}{5}\right)\).

\(2\ln\left(\frac{11}{5}\right)\approx 1.577\)

- Identify the derivative of the linear exponent. - Reverse the chain rule to find an antiderivative. - Recall that \(e^0=1\). - Track the negative sign when evaluating the limits.

1. An antiderivative of \(e^{1-0.5x}\) is \(-2e^{1-0.5x}\), because the derivative of \(1-0.5x\) is \(-0.5\). 2. Evaluate at the limits: \([-2e^{1-0.5x}]_0^2=-2e^0-(-2e^1)\). 3. Simplify to obtain \(2e-2\).

\(2e-2\approx 3.437\)

- In a), compare the numerator with the derivative of the denominator. - In b), compare one factor with the derivative of the exponent. - Change the limits when you substitute. - Recall the antiderivatives of \(\frac{1}{u}\) and \(e^u\).

1. For a), let \(u=x^3+1\), so \(du=3x^2\,dx\). The limits become \(u=2\) and \(u=9\). Therefore, \(\int_1^2\frac{3x^2}{x^3+1}\,dx=\int_2^9\frac{1}{u}\,du=\ln(9)-\ln(2)=\ln\left(\frac{9}{2}\right)\). 2. For b), let \(u=x^2-x\), so \(du=(2x-1)\,dx\). The limits become \(u=0\) and \(u=2\). Thus, \(\int_0^2(2x-1)e^{x^2-x}\,dx=\int_0^2e^u\,du=e^2-1\).

a) \(\ln\left(\frac{9}{2}\right)\) b) \(e^2-1\)

- In a), compare the numerator with the derivative of the denominator. - Rewrite the denominator in b) using a negative exponent. - Account for the inner derivative of the linear expression. - Check each antiderivative by differentiating.

1. For a), let \(u=x^2-5x+10\), so \(du=(2x-5)\,dx\). Then \([\ln|x^2-5x+10|]_1^3=\ln(4)-\ln(6)=\ln\left(\frac{2}{3}\right)\). 2. For b), rewrite the integrand as \((2x+2)^{-3}\). An antiderivative is \(-\frac{1}{4}(2x+2)^{-2}\). 3. Evaluate: \([-\frac{1}{4(2x+2)^2}]_0^1=-\frac{1}{64}+\frac{1}{16}=\frac{3}{64}\).

a) \(\ln\left(\frac{2}{3}\right)\) b) \(\frac{3}{64}\)

- Look for a factor that is the derivative of an inner expression. - Update the limits when changing variables. - Recall the derivative of \(e^x\). - Identify a logarithmic antiderivative pattern in b).

1. For a), let \(u=x^2\), so \(du=2x\,dx\). The limits become \(0\) and \(\pi\): \(\int_0^{\sqrt{\pi}}x\sin(x^2)\,dx=\frac{1}{2}\int_0^\pi\sin(u)\,du\). 2. Evaluate: \(\frac{1}{2}[-\cos(u)]_0^\pi=1\). 3. For b), let \(u=e^x+1\), so \(du=e^x\,dx\). The limits become \(2\) and \(e+1\): \(\int_0^1\frac{e^x}{e^x+1}\,dx=\int_2^{e+1}\frac{1}{u}\,du\). 4. Evaluate: \(\ln(e+1)-\ln(2)=\ln\left(\frac{e+1}{2}\right)\).

a) \(1\) b) \(\ln\left(\frac{e+1}{2}\right)\)

- Reverse the derivative rules for sine and cosine. - Account for each linear inner derivative. - Differentiate your result to check it.

1. For the cosine term, reverse the chain rule: \(\int 6\cos(3x-1)\,dx=2\sin(3x-1)\). 2. For the sine term, \(\int 2\sin(\frac{x}{2})\,dx=-4\cos(\frac{x}{2})\). 3. Therefore, \(F(x)=2\sin(3x-1)-4\cos\left(\frac{x}{2}\right)\).

\(F(x)=2\sin(3x-1)-4\cos\left(\frac{x}{2}\right)\)

- Rewrite the reciprocal square root using a negative exponent. - A reciprocal linear function leads to a logarithm. - Account for each linear inner derivative. - Differentiate your result to verify it.

1. For the first term, \(\int\frac{10}{2x+7}\,dx=5\ln(2x+7)\). The logarithm's argument is positive on the stated domain. 2. Rewrite the second term as \((4x+9)^{-\frac{1}{2}}\). Then \(\int(4x+9)^{-\frac{1}{2}}\,dx=\frac{1}{2}\sqrt{4x+9}\). 3. Thus \(F(x)=5\ln(2x+7)+\frac{1}{2}\sqrt{4x+9}\).

\(F(x)=5\ln(2x+7)+\frac{1}{2}\sqrt{4x+9}\)

- Identify the expression inside the power. - Reverse the chain rule for a linear inner function. - Differentiate your result to verify it.

1. The inner function is \(u(x)=\frac{1}{2}x+2\), and the outer function is \(v(u)=u^6\). 2. An antiderivative of \(u^6\) with respect to \(u\) is \(\frac{1}{7}u^7\). 3. Since \(u'(x)=\frac{1}{2}\), divide by \(\frac{1}{2}\): \(F(x)=\frac{2}{7}\left(\frac{1}{2}x+2\right)^7\).

Inner function: \(u(x)=\frac{1}{2}x+2\) Outer function: \(v(u)=u^6\) Antiderivative: \(F(x)=\frac{2}{7}\left(\frac{1}{2}x+2\right)^7\)

- Rewrite the reciprocal square root as a power. - Identify the derivative of the expression under the radical. - Reverse the power and chain rules.

1. Rewrite \(f(x)=(4-3x)^{-\frac{1}{2}}\). 2. The inner function is \(u(x)=4-3x\), and the outer function is \(v(u)=u^{-\frac{1}{2}}\). 3. An antiderivative of \(u^{-\frac{1}{2}}\) is \(2u^{\frac{1}{2}}\). Since \(u'(x)=-3\), divide by \(-3\): \(F(x)=-\frac{2}{3}\sqrt{4-3x}\).

Inner function: \(u(x)=4-3x\) Outer function: \(v(u)=u^{-\frac{1}{2}}\) Antiderivative: \(F(x)=-\frac{2}{3}\sqrt{4-3x}\)

- Identify the linear inner function in each expression. - Divide by the derivative of the inner function when reversing the chain rule. - Rewrite the rational power in b) using a negative exponent. - Differentiate each result to check it.

1. For a), reverse the power and chain rules: \(F(x)=\frac{1}{2}\left(\frac{2}{5}x-2\right)^5\). 2. For b), rewrite the integrand as \(3(1-2x)^{-2}\). An antiderivative is \(G(x)=\frac{3}{2}(1-2x)^{-1}=\frac{3}{2(1-2x)}\). 3. For c), account for the inner derivative \(4\): \(H(x)=\frac{1}{2}\sin\left(4x+\frac{\pi}{2}\right)\).

a) \(F(x)=\frac{1}{2}\left(\frac{2}{5}x-2\right)^5\) b) \(G(x)=\frac{3}{2(1-2x)}\) c) \(H(x)=\frac{1}{2}\sin\left(4x+\frac{\pi}{2}\right)\)

- Account for the derivative of the linear exponent. - Use the Fundamental Theorem after finding the antiderivative. - Recall that \(e^0=1\).

1. Reverse the chain rule to obtain \(F(x)=2e^{2x-1}\). 2. Apply the Fundamental Theorem: \(\int_{0.5}^{1}f(x)\,dx=[2e^{2x-1}]_{0.5}^{1}\). 3. Evaluate: \(2e^1-2e^0=2e-2\).

One antiderivative is \(F(x)=2e^{2x-1}\), and the integral equals \(2e-2\).

- Use the linear denominator as the substitution. - Transform both \(dx\) and the limits. - Recall the antiderivative of \(\frac{1}{u}\). - Simplify using logarithm properties.

1. Let \(u=2x+2\), so \(du=2\,dx\). The limits become \(u=2\) and \(u=8\). 2. Then \(\int_0^3\frac{4}{2x+2}\,dx=2\int_2^8\frac{1}{u}\,du\). 3. Evaluate: \(2[\ln|u|]_2^8=2\ln(4)=4\ln(2)\).

\(4\ln(2)\approx 2.773\)

- Track the negative sign when transforming \(dx\). - Reversing the new limits can remove that negative sign. - Recall the logarithmic antiderivative. - Combine the logarithm difference into one logarithm.

1. Let \(u=12-3x\), so \(du=-3\,dx\). The limits become \(u=15\) and \(u=6\). 2. Then \(\int_{-1}^{2}\frac{6}{12-3x}\,dx=-2\int_{15}^{6}\frac{1}{u}\,du=2\int_6^{15}\frac{1}{u}\,du\). 3. Evaluate: \(2[\ln|u|]_6^{15}=2\ln\left(\frac{15}{6}\right)=2\ln\left(\frac{5}{2}\right)\).

\(2\ln\left(\frac{5}{2}\right)\approx 1.833\)

- In a), account for the derivative of the linear exponent. - In b), compare the outside factor with the derivative of the exponent. - Think of the chain rule in reverse.

1. For a), reverse the chain rule. The derivative of \(4-0.5x\) is \(-0.5\), so \(F(x)=\frac{-2}{-0.5}e^{4-0.5x}=4e^{4-0.5x}\). 2. For b), the derivative of \(x^2-3x+1\) is \(2x-3\), and \(6x-9=3(2x-3)\). Therefore, \(F(x)=3e^{x^2-3x+1}\).

a) \(F(x)=4e^{4-0.5x}\) b) \(F(x)=3e^{x^2-3x+1}\)

- Compare the numerator of the first fraction with the derivative of its denominator. - Use the logarithmic antiderivative pattern. - Rewrite the second term with a negative exponent.

1. The derivative of \(x^2-3x+10\) is \(2x-3\), and \(4x-6=2(2x-3)\). Thus, \(\int\frac{4x-6}{x^2-3x+10}\,dx=2\ln(x^2-3x+10)\). The quadratic is always positive because its discriminant is \(9-40=-31\). 2. Rewrite \(\frac{1}{x^2}\) as \(x^{-2}\): \(\int x^{-2}\,dx=-x^{-1}=-\frac{1}{x}\). 3. Therefore, \(F(x)=2\ln(x^2-3x+10)-\frac{1}{x}\).

\(F(x)=2\ln(x^2-3x+10)-\frac{1}{x}\)

- Differentiate the denominator. - Look for the logarithmic antiderivative pattern. - Apply the bounds after finding an antiderivative.

1. The numerator is the derivative of the denominator, so \(F(x)=\ln(2x^2+1)\). 2. Apply the Fundamental Theorem: \(\int_0^1\frac{4x}{2x^2+1}\,dx=[\ln(2x^2+1)]_0^1\). 3. Evaluate: \(\ln(3)-\ln(1)=\ln(3)\).

One antiderivative is \(F(x)=\ln(2x^2+1)\). The integral equals \(\ln(3)\).

- Differentiate the denominator. - Adjust the constant factor to match the logarithmic pattern. - Simplify the expression at the upper bound before evaluating the logarithm. - Recall \(\ln(e)=1\) and \(\ln(1)=0\).

1. Since the derivative of \(x^2+1\) is \(2x\), an antiderivative is \(F(x)=\frac{1}{2}\ln(x^2+1)\). 2. Apply the bounds: \(\left[\frac{1}{2}\ln(x^2+1)\right]_0^{\sqrt{e-1}}\) \(=\frac{1}{2}\ln(e)-\frac{1}{2}\ln(1)=\frac{1}{2}\).

\(\frac{1}{2}\)

- Compare the factor outside the power with the derivative of the expression inside it. - Think of the chain rule in reverse. - Differentiate your proposed antiderivative to check it. - Apply the bounds only after finding an antiderivative.

1. The integrand has the form \(g'(x)(g(x))^3\), where \(g(x)=x^2-1\) and \(g'(x)=2x\). 2. Reversing the chain rule gives the antiderivative \(G(x)=\frac{1}{4}(x^2-1)^4\). 3. Apply the Fundamental Theorem: \(\left[\frac{1}{4}(x^2-1)^4\right]_1^2=\frac{1}{4}(3^4)-0=\frac{81}{4}\).

\(\frac{81}{4}=20.25\)

- Use the chain rule to differentiate the logarithm. - Compare the function in b) with your derivative from a). - Account for the constant factor when reversing differentiation.

1. By the chain rule, \(h'(x)=\frac{1}{x^2+5}\cdot 2x=\frac{2x}{x^2+5}\). 2. Since \(k(x)=\frac{1}{2}h'(x)\), one antiderivative is \(K(x)=\frac{1}{2}h(x)=\frac{1}{2}\ln(x^2+5)\).

a) \(h'(x)=\frac{2x}{x^2+5}\) b) \(K(x)=\frac{1}{2}\ln(x^2+5)\)

- Compare the numerator with the derivative of the denominator. - Use the logarithmic antiderivative pattern. - Evaluate the sine values at the bounds. - Combine the difference of logarithms.

1. The numerator is the derivative of \(\sin(x)+2\), so an antiderivative is \(F(x)=\ln(\sin(x)+2)\). The logarithm argument is positive on the interval. 2. Apply the bounds: \([\ln(\sin(x)+2)]_0^{\frac{\pi}{2}}=\ln(3)-\ln(2)\). 3. Use the quotient property of logarithms: \(\ln(3)-\ln(2)=\ln\left(\frac{3}{2}\right)\).

\(\ln\left(\frac{3}{2}\right)\)

- Differentiate the expression inside the power. - Rewrite the outside factor to match that derivative. - Apply the power rule after substituting.

1. Let \(u=2x^2-5\), so \(du=4x\,dx\) and \(x\,dx=\frac{1}{4}du\). 2. Then \(\int x(2x^2-5)^4\,dx=\frac{1}{4}\int u^4\,du\) \(=\frac{1}{20}u^5\). 3. Substitute back: \(F(x)=\frac{1}{20}(2x^2-5)^5\).

\(F(x)=\frac{1}{20}(2x^2-5)^5\)

- Differentiate the denominator. - Compare that derivative with the numerator. - Use the logarithmic antiderivative pattern. - Use the domain to determine whether absolute value is needed.

1. The derivative of \(x^3+8\) is \(3x^2\), so \(f(x)=\frac{1}{3}\cdot\frac{3x^2}{x^3+8}\). 2. Apply the logarithmic antiderivative pattern: \(F(x)=\frac{1}{3}\ln|x^3+8|\). 3. Since \(x>-2\), the expression \(x^3+8\) is positive. Therefore, \(F(x)=\frac{1}{3}\ln(x^3+8)\).

\(F(x)=\frac{1}{3}\ln(x^3+8)\)

- Compare the first factor with the derivative of the expression inside the power. - Reverse the chain rule or use substitution. - Apply the bounds after finding the antiderivative.

1. Let \(u=x^3+x+1\), so \(du=(3x^2+1)\,dx\). 2. Then \(\int(3x^2+1)(x^3+x+1)^2\,dx=\int u^2\,du=\frac{u^3}{3}\). Thus, \(F(x)=\frac{1}{3}(x^3+x+1)^3\). 3. Evaluate: \(F(1)-F(0)=\frac{1}{3}(3^3)-\frac{1}{3}(1^3)=\frac{26}{3}\).

One antiderivative is \(F(x)=\frac{1}{3}(x^3+x+1)^3\). The integral equals \(\frac{26}{3}\approx 8.67\).

- Compare the numerator with the derivative of the denominator. - Use the logarithmic antiderivative pattern. - Recall that \(e^{\ln(a)}=a\). - Combine the difference of logarithms.

1. The numerator is the derivative of the denominator, so \(F(x)=\ln(e^x+3)\). 2. Apply the bounds: \([\ln(e^x+3)]_0^{\ln(2)}=\ln(5)-\ln(4)\). 3. Combine the logarithms: \(\ln(5)-\ln(4)=\ln\left(\frac{5}{4}\right)\).

One antiderivative is \(F(x)=\ln(e^x+3)\). The integral equals \(\ln\left(\frac{5}{4}\right)\approx 0.223\).

- Differentiate the expression under the square root. - Use that expression as the substitution. - Rewrite the square root as a fractional power. - Substitute the original variable back at the end.

1. Let \(u=x^3+1\), so \(du=3x^2\,dx\) and \(x^2\,dx=\frac{1}{3}du\). 2. Then \(\int x^2\sqrt{x^3+1}\,dx=\frac{1}{3}\int u^{1/2}\,du\) \(=\frac{1}{3}\cdot\frac{2}{3}u^{3/2}\). 3. Substitute back: \(F(x)=\frac{2}{9}(x^3+1)^{3/2}\).

\(F(x)=\frac{2}{9}(x^3+1)^{3/2}\)

- Compare the outside factor with the derivative of the expression inside the power. - Use substitution and transform the bounds. - Check the antiderivative by differentiating with the chain rule. - Evaluate the transformed power integral.

1. Let \(u=x^3+1\), so \(du=3x^2\,dx\) and \(12x^2\,dx=4\,du\). The new bounds are \(u=1\) and \(u=2\). 2. Then \(\int_0^1 12x^2(x^3+1)^3\,dx=4\int_1^2u^3\,du\). 3. An antiderivative of the original integrand is \(F(x)=(x^3+1)^4\). 4. Evaluate: \([u^4]_1^2=2^4-1^4=15\).

One antiderivative is \(F(x)=(x^3+1)^4\). The integral equals \(15\).

- Compare the numerator with the derivative of the expression under the square root. - Rewrite the reciprocal square root as a negative power. - Either transform the bounds or substitute back before evaluating.

1. Let \(u=x^2+9\), so \(du=2x\,dx\). 2. Then \(\int\frac{x}{\sqrt{x^2+9}}\,dx=\frac{1}{2}\int u^{-1/2}\,du=\sqrt{u}\). Thus, \(F(x)=\sqrt{x^2+9}\). 3. Evaluate: \([\sqrt{x^2+9}]_0^4=5-3=2\).

One antiderivative is \(F(x)=\sqrt{x^2+9}\). The integral equals \(2\).

- Differentiate the denominator. - Compare that derivative with the numerator. - Use the logarithmic antiderivative pattern. - Combine the logarithms at the end.

1. The derivative of \(x^2+2\) is \(2x\), so an antiderivative is \(F(x)=\frac{1}{2}\ln(x^2+2)\). 2. Evaluate: \(\left[\frac{1}{2}\ln(x^2+2)\right]_0^2=\frac{1}{2}(\ln(6)-\ln(2))\). 3. Combine the logarithms: \(\frac{1}{2}\ln\left(\frac{6}{2}\right)=\frac{1}{2}\ln(3)\).

\(\frac{1}{2}\ln(3)\)

- Compare the factor outside the square root with the derivative of the expression inside it. - A substitution can simplify an integrand when one factor is the derivative of another expression. - Use the given point to determine the constant of integration. - Rewrite the square root as a power with exponent \(\frac{1}{2}\).

1. Use the substitution \(u=x^2+9\), so \(du=2x\,dx\). 2. Then \(\int 2x\sqrt{x^2+9}\,dx=\int u^{\frac{1}{2}}\,du=\frac{2}{3}u^{\frac{3}{2}}+C\). 3. Substitute back: \(F(x)=\frac{2}{3}(x^2+9)^{\frac{3}{2}}+C\). 4. Use \(F(0)=20\): \(\frac{2}{3}(9)^{\frac{3}{2}}+C=20\). Since \(9^{\frac{3}{2}}=27\), this gives \(18+C=20\), so \(C=2\). 5. Therefore, \(F(x)=\frac{2}{3}(x^2+9)^{\frac{3}{2}}+2\).

\(F(x)=\frac{2}{3}(x^2+9)^{\frac{3}{2}}+2\)

- Differentiate each proposed antiderivative. - For a linear inner function, compare multiplication by the inner derivative with division by it. - Consider why the antiderivative of \(\frac{g'(x)}{g(x)}\) is written with \(\ln|g(x)|\).

1. For A, differentiate the proposal: \(\frac{d}{dx}[-3\cos(3x)]=9\sin(3x)\), not \(\sin(3x)\). The inner derivative should be divided out. A correct antiderivative is \(F(x)=-\frac{1}{3}\cos(3x)\). 2. For B, \(\frac{d}{dx}[0.5e^{2x-4}]=0.5\cdot 2e^{2x-4}=e^{2x-4}\), so the proposal is correct. 3. For C, \(\frac{d}{dx}[\frac{1}{4}(4x+2)^4]=4(4x+2)^3\). The proposal does not divide by the inner derivative \(4\). A correct antiderivative is \(F(x)=\frac{1}{16}(4x+2)^4\). 4. For D, differentiating \(2\ln|0.5x+1|\) gives \(2\cdot\frac{0.5}{0.5x+1}=\frac{1}{0.5x+1}\). The absolute value makes the formula valid on intervals on either side of \(x=-2\). Thus a correct full-domain formula is \(F(x)=2\ln|0.5x+1|\).

a) Incorrect; the inner derivative was multiplied instead of divided. A correct antiderivative is \(F(x)=-\frac{1}{3}\cos(3x)\). b) Correct. c) Incorrect; the factor from the inner derivative was not divided out. A correct antiderivative is \(F(x)=\frac{1}{16}(4x+2)^4\). d) The proposed formula is not valid on intervals where \(0.5x+1<0\). A full-domain formula is \(F(x)=2\ln|0.5x+1|\).

- Differentiate each proposed antiderivative using the chain and product rules. - Look for the structure \(\frac{g'(x)}{g(x)}\). - Check whether a constant factor is needed to match the derivative exactly.

1. For A, \(\frac{d}{dx}\ln(x^2+1)=\frac{2x}{x^2+1}\), which is twice the given function. A correct antiderivative is \(F(x)=\frac{1}{2}\ln(x^2+1)\). 2. For B, \(\frac{d}{dx}e^{x^3}=3x^2e^{x^3}\), so the proposal is correct. 3. For C, the numerator is the derivative of the denominator. Therefore, \(\int\frac{e^x}{e^x+2}\,dx=\ln(e^x+2)+C\). The proposed product differentiates to additional terms, so it is incorrect. 4. For D, \(\frac{d}{dx}\sin^2(x)=2\sin(x)\cos(x)\), so the proposal is correct.

a) Incorrect; a factor of \(\frac{1}{2}\) is needed. A correct antiderivative is \(F(x)=\frac{1}{2}\ln(x^2+1)\). b) Correct. c) Incorrect; the numerator is already the derivative of the denominator. A correct antiderivative is \(F(x)=\ln(e^x+2)\). d) Correct.

- Split the integrand into two simpler terms. - In the rational term, compare the numerator with the derivative of the denominator. - For \((3x-1)^2\), account for the derivative of the linear inner function. - Evaluate each part before adding the results.

1. Split the integral into two parts: \(\int_0^2\frac{2x}{x^2+1}\,dx+\int_0^2(3x-1)^2\,dx\). 2. For the first part, use \(u=x^2+1\), so \(du=2x\,dx\). Then \(\int_0^2\frac{2x}{x^2+1}\,dx=[\ln(x^2+1)]_0^2=\ln(5)\). 3. For the second part, an antiderivative is \(\frac{1}{9}(3x-1)^3\). Therefore, \([\frac{1}{9}(3x-1)^3]_0^2=\frac{1}{9}[5^3-(-1)^3]=14\). 4. Add the results to obtain \(14+\ln(5)\).

\(14+\ln(5)\)

- In a), compare the derivative of the exponent with the factor outside the exponential. - Integrate the two terms in b) separately. - When integrating \(\cos(kx)\), account for the constant \(k\). - Substitute the limits carefully and check signs of trigonometric values.

1. For a), let \(u=-\frac{1}{2}x^2\), so \(du=-x\,dx\). Then an antiderivative is \(-e^{-\frac{1}{2}x^2}\). 2. Evaluate: \([-e^{-\frac{1}{2}x^2}]_0^2=-e^{-2}+1=1-e^{-2}\). 3. For b), integrate term by term. An antiderivative is \(-\frac{1}{2}\cos(x)+2\sin\left(\frac{x}{2}\right)\). 4. Evaluate: \([ -\frac{1}{2}\cos(x)+2\sin(\frac{x}{2}) ]_0^\pi=(\frac{1}{2}+2)-(-\frac{1}{2})=3\).

a) \(1-e^{-2}\approx 0.865\) b) \(3\)

- How do you integrate a power of a linear expression? - Differentiate your antiderivative to check the chain-rule factor. - Substitute the given input and output to determine the constant.

1. Use substitution for the linear expression: \(F(t)=-\frac{1}{2}(1-2t)^4+C\). 2. Apply the condition \(F(1)=5\): \(-\frac{1}{2}(1-2)^4+C=5\). 3. Since \((1-2)^4=1\), \(-\frac{1}{2}+C=5\), so \(C=\frac{11}{2}\). 4. Thus, \(F(t)=-\frac{1}{2}(1-2t)^4+\frac{11}{2}\).

\(F(t)=-\frac{1}{2}(1-2t)^4+\frac{11}{2}\)

- Rewrite the rational expression in 1 as a sum of powers of \(x\). - In 2, compare the numerator with the derivative of the denominator. - In 3, account for the derivative of \(0.5x\). - For 4 and 5, ask whether any of the specifically listed antiderivative patterns applies.

1. For 1, use the power, sum, and constant multiple rules. Rewrite the integrand as \(1+x^{-2}\). Then \([x-\frac{1}{x}]_1^2=\frac{3}{2}\). 2. For 2, use the logarithmic rule with \(g(x)=x^3+1\), since \(g'(x)=3x^2\). Thus \([\ln(x^3+1)]_0^1=\ln(2)\). 3. For 3, use a linear substitution, or account directly for the inner derivative. An antiderivative is \(2e^{0.5x}\), so \([2e^{0.5x}]_0^2=2e-2\). 4. The fourth integral cannot be evaluated using only the listed methods; a standard method is integration by parts. 5. The fifth integral cannot be evaluated using only the listed methods; its antiderivative involves \(\arctan(x)\).

1. Power, sum, and constant multiple rules: \(\frac{3}{2}\) 2. Logarithmic rule: \(\ln(2)\) 3. Linear substitution: \(2e-2\) 4. Not evaluable using only the listed methods. 5. Not evaluable using only the listed methods.

- Expand the product in 1 before integrating. - In 2, use the antiderivative pattern for the reciprocal of a linear function. - In 4, compare the numerator with the derivative of the denominator. - Decide whether the listed rules include a method for products such as \(x\cos(x)\).

1. For 1, use the power rule after expanding: \((x-1)(x+1)=x^2-1\). Then \([\frac{1}{3}x^3-x]_0^2=\frac{2}{3}\). 2. For 2, use a linear substitution, or the reciprocal-of-a-linear-function pattern. An antiderivative is \(\frac{1}{2}\ln|2x-3|\), so \([\frac{1}{2}\ln|2x-3|]_2^5=\frac{1}{2}\ln(7)\). 3. The third integral cannot be evaluated using only the listed methods; a standard method is integration by parts. 4. For 4, use the logarithmic rule because the numerator is the derivative of the denominator. Therefore, \([\ln(x^2+3x)]_1^2=\ln(10)-\ln(4)=\ln(\frac{5}{2})\). 5. The fifth integral cannot be evaluated using only the listed methods; it requires a more advanced substitution or a known special antiderivative.

1. Power rule after expanding: \(\frac{2}{3}\) 2. Linear substitution: \(\frac{1}{2}\ln(7)\) 3. Not evaluable using only the listed methods. 4. Logarithmic rule: \(\ln(\frac{5}{2})\) 5. Not evaluable using only the listed methods.

- Rewrite the reciprocal square root using a negative exponent. - Use the integration rule for \((ax+b)^n\). - The point gives a specific value of the antiderivative. - Substitute the constant back into the final expression.

1. Rewrite the integrand as \(f(x)=6(2x+9)^{-\frac{1}{2}}\). 2. Integrate using substitution: \(F(x)=6\sqrt{2x+9}+C\). 3. Use \(F(0)=10\): \(6\sqrt{9}+C=10\). 4. Since \(18+C=10\), \(C=-8\). Therefore, \(F(x)=6\sqrt{2x+9}-8\).

\(F(x)=6\sqrt{2x+9}-8\)

- Rewrite the square root as a rational power. - Use a substitution for the linear expression inside the radical. - Find an antiderivative before evaluating at the limits. - Be careful with the sign of the lower-limit value.

1. To integrate the radical term, let \(u=2x+4\), so \(du=2\,dx\) and \(dx=\frac{1}{2}\,du\). 2. Then \(\int\sqrt{2x+4}\,dx=\frac{1}{2}\int u^{1/2}\,du=\frac{1}{3}u^{3/2}+C\). Thus, an antiderivative of \(f\) is \(F(x)=4x-\frac{1}{3}(2x+4)^{3/2}\). 3. Evaluate: \(F(0)=-\frac{8}{3}\) and \(F(-2)=-8\). 4. Therefore, \(F(0)-F(-2)=-\frac{8}{3}+8=\frac{16}{3}\).

\(\frac{16}{3}\)

- Rewrite the radical as a rational power. - Use a substitution for the linear expression inside the radical. - Account for the coefficient of \(x\) in the substitution. - Simplify the expression inside the radical at each limit before evaluating the power.

1. Let \(u=4-\frac{1}{2}x\). Then \(du=-\frac{1}{2}\,dx\), so \(dx=-2\,du\). 2. An antiderivative is \(-2\int u^{1/2}\,du=-\frac{4}{3}u^{3/2}+C\). Therefore, \(G(x)=-\frac{4}{3}\left(4-\frac{1}{2}x\right)^{3/2}\). 3. Evaluate: \(G(8)=0\) and \(G(0)=-\frac{4}{3}\cdot4^{3/2}=-\frac{32}{3}\). 4. Thus, \(G(8)-G(0)=\frac{32}{3}\).

\(\frac{32}{3}\)

- What happens when the numerator and denominator are identical? - For the logarithmic rule, how must the numerator relate to the denominator? - Which functions have derivative \(e^x\)?

1. For a), choose the denominator \(e^x\). Then the integrand is \(1\), so \(\int_0^1 1\,dx=1\). 2. For b), choose \(e^x+1\). Its derivative is the numerator \(e^x\), so \(\int_0^1\frac{e^x}{e^x+1}\,dx=[\ln(e^x+1)]_0^1\). 3. Evaluate to obtain \(\ln(e+1)-\ln(2)=\ln\left(\frac{e+1}{2}\right)\).

a) Denominator \(e^x\); value \(1\) b) One possible denominator is \(e^x+1\); value \(\ln\left(\frac{e+1}{2}\right)\)

- Work with the two terms separately. - Use the quotient rule for the rational term when differentiating. - Compare the rational numerator with the derivative of its denominator when integrating. - Apply the chain rule and its reverse to the trigonometric term.

1. Differentiate the rational term using the quotient rule: \(\frac{d}{dx}\left(\frac{4x}{2x^2+3}\right)=\frac{4(2x^2+3)-4x(4x)}{(2x^2+3)^2}=\frac{12-8x^2}{(2x^2+3)^2}\). 2. Differentiate the sine term using the chain rule: \(\frac{d}{dx}\sin(2x+1)=2\cos(2x+1)\). Thus \(f'(x)=\frac{12-8x^2}{(2x^2+3)^2}+2\cos(2x+1)\). 3. For an antiderivative, the numerator \(4x\) is the derivative of \(2x^2+3\), so the first term integrates to \(\ln(2x^2+3)\). 4. The sine term integrates to \(-\frac{1}{2}\cos(2x+1)\). Therefore, \(F(x)=\ln(2x^2+3)-\frac{1}{2}\cos(2x+1)\).

\(f'(x)=\frac{12-8x^2}{(2x^2+3)^2}+2\cos(2x+1)\) \(F(x)=\ln(2x^2+3)-\frac{1}{2}\cos(2x+1)\)

- Rewrite the reciprocal square root as a negative power. - First find the general antiderivative with \(C\). - Use \(F(0)=5\) to determine \(C\). - Differentiate your result to verify it.

1. Rewrite \(f(x)=(2x+9)^{-\frac{1}{2}}\). 2. Reverse the power and chain rules: \(F(x)=\sqrt{2x+9}+C\). 3. Use the condition \(F(0)=5\): \(3+C=5\), so \(C=2\). 4. Therefore, \(F(x)=\sqrt{2x+9}+2\).

\(F(x)=\sqrt{2x+9}+2\)

- Use the binomial theorem or repeated multiplication to expand. - For substitution, identify the derivative of the linear inner function. - Update the limits after changing variables. - Compare the exact values from both approaches.

1. Expand: \(\left(\frac{1}{2}x+1\right)^3=\frac{1}{8}x^3+\frac{3}{4}x^2+\frac{3}{2}x+1\). 2. Integrate term by term: \([\frac{1}{32}x^4+\frac{1}{4}x^3+\frac{3}{4}x^2+x]_0^2=\frac{15}{2}\). 3. For substitution, let \(u=\frac{1}{2}x+1\), so \(du=\frac{1}{2}\,dx\). The limits become \(1\) and \(2\): \(2\int_1^2u^3\,du=2[\frac{1}{4}u^4]_1^2=\frac{15}{2}\). 4. Both methods give the same value.

\(\frac{15}{2}\) by both methods

- Use the binomial coefficients for a fourth power. - Track the negative derivative of \(1-2x\). - Update the limits after substitution. - Compare the exact results from both methods.

1. Expand: \((1-2x)^4=1-8x+24x^2-32x^3+16x^4\). 2. Integrate term by term: \([x-4x^2+8x^3-8x^4+\frac{16}{5}x^5]_0^1=\frac{1}{5}\). 3. For substitution, let \(u=1-2x\), so \(du=-2\,dx\). The limits become \(1\) and \(-1\): \(-\frac{1}{2}\int_1^{-1}u^4\,du=\frac{1}{2}\int_{-1}^{1}u^4\,du=\frac{1}{5}\). 4. Both methods give the same value.

\(\frac{1}{5}\) by both methods

- Split the integral using linearity. - Reverse the power and chain rules for each linear inner function. - Rewrite the reciprocal square using a negative exponent. - Preserve the subtraction between the two terms.

1. An antiderivative of \((2x+1)^3\) is \(\frac{1}{8}(2x+1)^4\). Its contribution is \([\frac{1}{8}(2x+1)^4]_0^1=10\). 2. An antiderivative of \(\frac{3}{(3x+1)^2}\) is \(-\frac{1}{3x+1}\). Its contribution is \([-\frac{1}{3x+1}]_0^1=\frac{3}{4}\). 3. The second term is subtracted in the original integrand, so the total is \(10-\frac{3}{4}=\frac{37}{4}\).

\(\frac{37}{4}\)

- Treat the two terms separately. - Account for the inner derivative in the sine term. - Rewrite the reciprocal square root as a negative power. - Use the cosine values at \(0\) and \(\pi\).

1. An antiderivative of \(\sin(\frac{\pi x}{4})\) is \(-\frac{4}{\pi}\cos(\frac{\pi x}{4})\). Its contribution is \([ -\frac{4}{\pi}\cos(\frac{\pi x}{4}) ]_0^4=\frac{8}{\pi}\). 2. Rewrite the second term as \((2x+1)^{-\frac{1}{2}}\). An antiderivative is \(\sqrt{2x+1}\), so its contribution is \([\sqrt{2x+1}]_0^4=3-1=2\). 3. Add the results to obtain \(\frac{8}{\pi}+2\).

\(\frac{8}{\pi}+2\)

- Factor the quadratic as a perfect square. - Simplify the power of a power. - Reverse the power and chain rules. - Track the sign at the lower limit.

1. Recognize the perfect square: \(4x^2-4x+1=(2x-1)^2\). 2. Therefore, the integrand is \((2x-1)^6\). 3. An antiderivative is \(\frac{1}{14}(2x-1)^7\). 4. Evaluate: \([\frac{1}{14}(2x-1)^7]_0^1=\frac{1}{14}-(-\frac{1}{14})=\frac{1}{7}\).

\(\frac{1}{7}\)

- Expand the square in a) before integrating. - In b), compare the numerator with the derivative of the denominator. - Recall the logarithmic antiderivative pattern.

1. For a), first expand: \((e^{2x}+1)^2=e^{4x}+2e^{2x}+1\). Integrate term by term: \(F(x)=\frac{1}{4}e^{4x}+e^{2x}+x\). 2. For b), let \(u=e^{2x}+3\). Then \(du=2e^{2x}\,dx\), so \(\int\frac{4e^{2x}}{e^{2x}+3}\,dx=2\int\frac{1}{u}\,du=2\ln|u|\). Since \(e^{2x}+3>0\), \(F(x)=2\ln(e^{2x}+3)\).

a) \(F(x)=\frac{1}{4}e^{4x}+e^{2x}+x\) b) \(F(x)=2\ln(e^{2x}+3)\)

- Rewrite the first fraction as a constant plus a simpler fraction. - Compare the numerator of the second fraction with the derivative of its denominator. - Adjust the constant factor before using the logarithmic pattern.

1. Rewrite the first fraction: \(\frac{x+3}{x+2}=\frac{x+2+1}{x+2}=1+\frac{1}{x+2}\). Therefore, \(\int\frac{x+3}{x+2}\,dx=x+\ln|x+2|\). 2. For the second fraction, the derivative of \(2e^x+4\) is \(2e^x\). Hence, \(\int\frac{e^x}{2e^x+4}\,dx=\frac{1}{2}\ln(2e^x+4)\). 3. Combine the results: \(F(x)=x+\ln|x+2|-\frac{1}{2}\ln(2e^x+4)\).

\(F(x)=x+\ln|x+2|-\frac{1}{2}\ln(2e^x+4)\)

- Look for an inner function and its derivative. - For a product, check whether one factor is the derivative of the expression inside a power. - For a fraction, compare the numerator with the derivative of the denominator. - In d), rewrite \(x\) in terms of \(x+1\).

1. For a), reverse the chain rule: \(F(x)=(2x+7)^4\). 2. For b), the first factor is the derivative of \(x^3-6x+1\), so \(F(x)=\frac{1}{6}(x^3-6x+1)^6\). 3. For c), the numerator is the derivative of the denominator, so \(F(x)=\ln|x^2+5x-2|\). 4. For d), write \(x=(x+1)-1\): \(x(x+1)^4=(x+1)^5-(x+1)^4\). Therefore, \(F(x)=\frac{1}{6}(x+1)^6-\frac{1}{5}(x+1)^5\).

a) \(F(x)=(2x+7)^4\) b) \(F(x)=\frac{1}{6}(x^3-6x+1)^6\) c) \(F(x)=\ln|x^2+5x-2|\) d) \(F(x)=\frac{1}{6}(x+1)^6-\frac{1}{5}(x+1)^5\)

- Account for the derivative of each exponent. - Reverse the sine and cosine derivative rules carefully. - Rewrite the fraction in d) with a negative power. - Differentiate each result to check it.

1. For a), account for the derivative \(-2\) of the exponent: \(F(x)=3e^{3-2x}\). 2. For b), the derivative of \(x^2+4x\) is \(2(x+2)\), so \(F(x)=\frac{1}{2}e^{x^2+4x}\). 3. For c), reverse the chain rule: \(F(x)=-\cos(5x-\pi)\). 4. For d), use \(u=x^2+4\), so \(du=2x\,dx\): \(\int\frac{6x}{(x^2+4)^2}\,dx=3\int u^{-2}\,du=-3u^{-1}\). Thus, \(F(x)=-\frac{3}{x^2+4}\).

a) \(F(x)=3e^{3-2x}\) b) \(F(x)=\frac{1}{2}e^{x^2+4x}\) c) \(F(x)=-\cos(5x-\pi)\) d) \(F(x)=-\frac{3}{x^2+4}\)

- Use the product and chain rules on the exponential term when differentiating. - Rewrite the square root as a power. - For integration, compare \(x\) with the derivative of \(-x^2\). - Reverse the power and chain rules for the linear expression under the square root.

1. For \(x>\frac{2}{3}\), differentiate the first term using the product and chain rules: \(\frac{d}{dx}[xe^{-x^2}]=e^{-x^2}-2x^2e^{-x^2}=(1-2x^2)e^{-x^2}\). 2. Differentiate the square-root term: \(\frac{d}{dx}(3x-2)^{\frac{1}{2}}=\frac{3}{2\sqrt{3x-2}}\). Thus \(f'(x)=(1-2x^2)e^{-x^2}+\frac{3}{2\sqrt{3x-2}}\) for \(x>\frac{2}{3}\). 3. For an antiderivative of the first term, use \(u=-x^2\), giving \(-\frac{1}{2}e^{-x^2}\). 4. For the second term, reverse the power and chain rules: \(\int(3x-2)^{\frac{1}{2}}\,dx=\frac{2}{9}(3x-2)^{\frac{3}{2}}\). 5. Therefore, \(F(x)=-\frac{1}{2}e^{-x^2}+\frac{2}{9}(3x-2)^{\frac{3}{2}}\).

\(f'(x)=(1-2x^2)e^{-x^2}+\frac{3}{2\sqrt{3x-2}}\) for \(x>\frac{2}{3}\) \(F(x)=-\frac{1}{2}e^{-x^2}+\frac{2}{9}(3x-2)^{\frac{3}{2}}\) for \(x\ge\frac{2}{3}\)