Free math worksheets

- Prefer to differentiate a factor that becomes simpler. - Choose \(dv\) so its antiderivative is immediate. - A single logarithmic function can be treated as a product with \(1\).

1. a) Choose \(u=x\) and \(dv=e^x\,dx\). Differentiating \(x\) simplifies it, while \(e^x\) is easy to integrate. 2. b) Choose \(u=x^2\) and \(dv=\cos(x)\,dx\). Differentiating the polynomial lowers its degree. 3. c) View the integrand as \(\ln(x)\cdot1\). Choose \(u=\ln(x)\) and \(dv=dx\), because differentiating the logarithm produces \(1/x\).

a) \(u=x\), \(dv=e^x\,dx\) b) \(u=x^2\), \(dv=\cos(x)\,dx\) c) \(u=\ln(x)\), \(dv=dx\)

- Use integration by parts for the product of a linear function and an exponential function. - Differentiate the linear factor and integrate the exponential factor. - Account for the inner derivative in \(e^{3x}\). - Factor out the exponential term when simplifying.

1. Use integration by parts with \(u=x+1\) and \(dv=e^{3x}\,dx\). Then \(du=dx\) and \(v=\frac{1}{3}e^{3x}\). 2. Therefore, \(\int(x+1)e^{3x}\,dx=\frac{1}{3}(x+1)e^{3x}-\frac{1}{3}\int e^{3x}\,dx\). 3. Evaluate the remaining integral: \(\frac{1}{3}(x+1)e^{3x}-\frac{1}{9}e^{3x}\). 4. Simplify to obtain \(F(x)=\left(\frac{x}{3}+\frac{2}{9}\right)e^{3x}\).

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

- Choose the factor that becomes simpler when differentiated. - Account for the inner derivative when integrating \(\cos(2x)\). - Track the sign when integrating sine. - Differentiate your result to verify it.

1. Use integration by parts with \(u=4x\) and \(dv=\cos(2x)\,dx\). Then \(du=4\,dx\) and \(v=\frac{1}{2}\sin(2x)\). 2. Apply the formula: \(\int 4x\cos(2x)\,dx=2x\sin(2x)-2\int\sin(2x)\,dx\). 3. Since \(\int 2\sin(2x)\,dx=-\cos(2x)\), the result is \(F(x)=2x\sin(2x)+\cos(2x)\).

\(F(x)=2x\sin(2x)+\cos(2x)\)

- Differentiate the linear factor. - Use integration by parts. - Account for the inner factor when integrating \(\sin(2x)\) and \(\cos(2x)\). - Track the signs carefully.

1. Use integration by parts with \(u=2-x\) and \(dv=\sin(2x)\,dx\). Then \(du=-dx\) and \(v=-\frac{1}{2}\cos(2x)\). 2. Therefore, \(F(x)=-\frac{1}{2}(2-x)\cos(2x)-\frac{1}{2}\int\cos(2x)\,dx\). 3. Since \(\int\cos(2x)\,dx=\frac{1}{2}\sin(2x)\), \(F(x)=\left(\frac{x}{2}-1\right)\cos(2x)-\frac{1}{4}\sin(2x)\).

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

- Choose the factor whose derivative becomes simpler. - Choose the other factor so its antiderivative is immediate. - After applying integration by parts, simplify the remaining power integral.

1. Choose \(u=\ln(x)\) and \(\text{d}v=x^3\,\text{d}x\), so \(\text{d}u=\frac{1}{x}\,\text{d}x\) and \(v=\frac{x^4}{4}\). 2. Integration by parts gives \(\frac{x^4}{4}\ln(x)-\frac{1}{4}\int x^3\,\text{d}x\). 3. The result is \(\frac{x^4}{4}\ln(x)-\frac{x^4}{16}+C\). 4. Differentiating \(\ln(x)\) simplifies it to \(1/x\), while attempting to integrate it would not simplify the product.

\(\frac{x^4}{4}\ln(x)-\frac{x^4}{16}+C\)

- Differentiate the algebraic factor and integrate the trigonometric factor. - Track the negative sign in the antiderivative of \(\sin(x)\). - Verify the final expression with the product rule.

1. Let \(u=x\) and \(\text{d}v=\sin(x)\,\text{d}x\), so \(\text{d}u=\text{d}x\) and \(v=-\cos(x)\). 2. Integration by parts gives \(-x\cos(x)+\int\cos(x)\,\text{d}x\). 3. Therefore, the antiderivative is \(-x\cos(x)+\sin(x)+C\). 4. Differentiation gives \(-\cos(x)+x\sin(x)+\cos(x)=x\sin(x)\).

\(-x\cos(x)+\sin(x)+C\)

- Differentiate the algebraic factor and integrate the exponential factor. - Track the negative sign in an antiderivative of \(e^{-x}\). - Apply the bounds to both terms before combining them.

1. Let \(u=x\) and \(\text{d}v=e^{-x}\,\text{d}x\). Then \(\text{d}u=\text{d}x\) and \(v=-e^{-x}\). 2. Integration by parts gives \(\int_0^1xe^{-x}\,\text{d}x=[-xe^{-x}]_0^1+\int_0^1e^{-x}\,\text{d}x\). 3. Evaluate: \(-\frac{1}{e}+[-e^{-x}]_0^1=-\frac{1}{e}+1-\frac{1}{e}=1-\frac{2}{e}\).

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

- Check the proposed result by differentiation before trying to repair it. - Think of the integrand as the product of a function and \(1\). - What term would cancel the extra derivative?

1. Differentiating Eleni's expression gives \(\ln(2x)+1\), so it has an extra constant term. 2. Treat \(\ln(2x)\) as a product with \(1\). The transformed integral is \(x\ln(2x)-\int1\,\text{d}x\). 3. The correct antiderivative is \(x\ln(2x)-x+C\).

Eleni's expression differentiates to \(\ln(2x)+1\), not \(\ln(2x)\). The correct antiderivative is \(x\ln(2x)-x+C\).

- Integrate the product-rule equality term by term. - Isolate the integral containing \(u(x)v'(x)\). - In the example, choose the factor that becomes simpler when differentiated.

1. Integrating the product-rule identity gives \(u(x)v(x)=\int u'(x)v(x)\,\text{d}x+\int u(x)v'(x)\,\text{d}x\). 2. Rearranging gives \(\int u(x)v'(x)\,\text{d}x=u(x)v(x)-\int v(x)u'(x)\,\text{d}x\). 3. For \(\int x e^x\,\text{d}x\), take \(u=x\) and \(v'=e^x\). Then \(u'=1\) and \(v=e^x\). 4. Therefore \(\int x e^x\,\text{d}x=xe^x-\int e^x\,\text{d}x\).

\(\int u(x)v'(x)\,\text{d}x=u(x)v(x)-\int v(x)u'(x)\,\text{d}x\). For \(\int x e^x\,\text{d}x\): \(xe^x-\int e^x\,\text{d}x\).

- Choose the factor that becomes simpler when differentiated. - Track the negative sign when integrating sine. - Recall the sine and cosine values at \(0\) and \(\frac{\pi}{2}\). - Evaluate the complete integration-by-parts expression at both limits.

1. Choose \(u=2x-1\) and \(dv=\sin(x)\,dx\). Then \(du=2\,dx\) and \(v=-\cos(x)\). 2. Apply integration by parts: \(\int_0^{\frac{\pi}{2}}(2x-1)\sin(x)\,dx=[-(2x-1)\cos(x)]_0^{\frac{\pi}{2}}+2\int_0^{\frac{\pi}{2}}\cos(x)\,dx\). 3. Combine the terms: \([-(2x-1)\cos(x)+2\sin(x)]_0^{\frac{\pi}{2}}\). 4. Substitute the limits. The upper value is \(2\), and the lower value is \(1\), so the integral equals \(1\).

\(1\)

- Choose one exponent whose exponential factor remains simple under differentiation. - For substitution, look for an exponent whose derivative is a constant multiple of \(x\). - Integration by parts is useful when differentiating one factor simplifies it. - Check that each chosen exponent leads to the method requested.

1. Choose the exponent \(x\). Then \(\int_0^1xe^x\,dx\) is evaluated by integration by parts. With \(u=x\) and \(dv=e^x\,dx\), an antiderivative is \((x-1)e^x\). Thus \([(x-1)e^x]_0^1=1\). 2. Choose the exponent \(x^2\). Then \(\int_0^1xe^{x^2}\,dx\) is evaluated by substitution. Let \(u=x^2\), so \(du=2x\,dx\). Therefore, \(\int_0^1xe^{x^2}\,dx=\frac{1}{2}\int_0^1e^u\,du=\frac{e-1}{2}\).

One valid pair is: Exponent \(x\): integration by parts gives \(1\). Exponent \(x^2\): substitution gives \(\frac{e-1}{2}\).

- Write the integrand as a product with \(1\). - Differentiate the logarithmic factor and integrate the constant factor. - Simplify the remaining integral after applying integration by parts. - Use \(\ln(1)=0\) at the lower limit.

1. Use integration by parts with \(u=\ln(5t)\) and \(dv=dt\). Then \(du=\frac{1}{t}\,dt\) and \(v=t\). 2. Thus \(\int\ln(5t)\,dt=t\ln(5t)-\int 1\,dt=t\ln(5t)-t\). 3. Evaluate the accumulation function: \(I_{0.2}(x)=[t\ln(5t)-t]_{0.2}^{x}\). 4. Since \(\ln(5\cdot 0.2)=\ln(1)=0\), \(I_{0.2}(x)=x\ln(5x)-x+0.2\).

\(I_{0.2}(x)=x\ln(5x)-x+0.2\)

- Use the specified choices to find both \(\text{d}u\) and \(v\). - Keep the definite-integral bounds when applying integration by parts. - After the repeated integral appears, treat \(I\) as an algebraic unknown.

1. With \(u=\ln x\) and \(\text{d}v=\frac{1}{x}\,\text{d}x\), we have \(\text{d}u=\frac{1}{x}\,\text{d}x\) and \(v=\ln x\). 2. Integration by parts gives \(I=[(\ln x)^2]_1^e-I\). 3. Since \([(\ln x)^2]_1^e=1\), the equation is \(I=1-I\). 4. Therefore \(2I=1\) and \(I=\frac{1}{2}\).

Integration by parts gives \(I=[(\ln x)^2]_1^e-I=1-I\), so \(I=\frac{1}{2}\).

- Differentiate the polynomial factor so its degree decreases. - Use integration by parts on the remaining product again. - Account for the factor \(2\) when integrating \(e^{2x}\). - Evaluate the complete antiderivative at the limits.

1. Choose \(u=x^2\) and \(dv=e^{2x}\,dx\). Then \(du=2x\,dx\) and \(v=\frac{1}{2}e^{2x}\), so \(\int_0^1x^2e^{2x}\,dx=[\frac{1}{2}x^2e^{2x}]_0^1-\int_0^1xe^{2x}\,dx\). 2. For the remaining integral, use \(u=x\) and \(dv=e^{2x}\,dx\): \(\int xe^{2x}\,dx=\frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x}\). 3. An antiderivative of the original integrand is \(F(x)=\left(\frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{4}\right)e^{2x}\). 4. Evaluate: \(F(1)-F(0)=\frac{1}{4}e^2-\frac{1}{4}=\frac{e^2-1}{4}\).

\(\frac{e^2-1}{4}\)

- Differentiate the polynomial factor to lower its degree. - Apply integration by parts again to the remaining product. - Track the signs when integrating sine. - Use the sine and cosine values at \(0\) and \(\frac{\pi}{2}\).

1. Choose \(u=x^2\) and \(dv=\cos(x)\,dx\). Then \(du=2x\,dx\) and \(v=\sin(x)\): \(\int_0^{\frac{\pi}{2}}x^2\cos(x)\,dx=[x^2\sin(x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}2x\sin(x)\,dx\). 2. For the remaining integral, choose \(u=2x\) and \(dv=\sin(x)\,dx\). Then \(du=2\,dx\) and \(v=-\cos(x)\), so \(\int 2x\sin(x)\,dx=-2x\cos(x)+2\sin(x)\). 3. An antiderivative of the original integrand is \(F(x)=x^2\sin(x)+2x\cos(x)-2\sin(x)\). 4. Evaluate: \(F(\frac{\pi}{2})-F(0)=\frac{\pi^2}{4}-2\).

\(\frac{\pi^2}{4}-2\)

- Choose the factor that becomes simpler when differentiated. - Use the integration-by-parts formula. - Recall the values of \(\ln(e)\) and \(\ln(1)\). - Combine like terms at the end.

1. Let \(u=\ln(x)\) and \(dv=x^2\,dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=\frac{x^3}{3}\). 2. Integration by parts gives \(\int x^2\ln(x)\,dx=\frac{x^3}{3}\ln(x)-\frac{1}{3}\int x^2\,dx\) \(=\frac{x^3}{3}\ln(x)-\frac{x^3}{9}\). 3. Evaluate from \(1\) to \(e\): \(\left[\frac{x^3}{3}\ln(x)-\frac{x^3}{9}\right]_1^e=\left(\frac{e^3}{3}-\frac{e^3}{9}\right)-\left(0-\frac{1}{9}\right)\) \(=\frac{2e^3+1}{9}\).

\(\frac{2e^3+1}{9}\)

- Differentiate the linear factor. - Be careful when finding an antiderivative of \(e^{-x}\). - Apply the bounds to each term carefully. - Simplify the exponential values at the end.

1. Let \(u=2x+1\) and \(dv=e^{-x}\,dx\). Then \(du=2\,dx\) and \(v=-e^{-x}\). 2. Apply integration by parts: \(\int_0^1(2x+1)e^{-x}\,dx=[-(2x+1)e^{-x}]_0^1+2\int_0^1e^{-x}\,dx\). 3. Evaluate: \([-(2x+1)e^{-x}]_0^1+[-2e^{-x}]_0^1\) \(=\left(-\frac{3}{e}+1\right)+\left(-\frac{2}{e}+2\right)=3-\frac{5}{e}\).

\(3-\frac{5}{e}\approx 1.1606\)

- Differentiate the factor that becomes simpler. - Compare what differentiation and integration do to a polynomial degree. - The goal is to make the remaining integral easier than the original.

1. For a), if \(u=e^{0.5x}\) and \(dv=(2x+4)\,dx\), then \(du=0.5e^{0.5x}\,dx\) and \(v=x^2+4x\). The new integral contains the quadratic factor \(x^2+4x\), so the polynomial degree increases from \(1\) to \(2\). 2. For b), choose \(u=2x+4\) and \(dv=e^{0.5x}\,dx\). Then \(du=2\,dx\) and \(v=2e^{0.5x}\). 3. Apply integration by parts: \(I=[(2x+4)2e^{0.5x}]_0^1-4\int_0^1e^{0.5x}\,dx\). 4. Evaluate: \(I=[4(x+2)e^{0.5x}]_0^1-[8e^{0.5x}]_0^1\) \(=(12\sqrt e-8)-(8\sqrt e-8)=4\sqrt e\).

a) The polynomial degree increases from \(1\) to \(2\). b) \(I=4\sqrt e\)

- Differentiate the logarithmic factor. - Apply integration by parts. - Use \(\ln(e)=1\) and \(\ln(1)=0\).

1. Let \(u=\ln(x)\) and \(dv=x\,dx\). Then \(du=\frac{1}{x}\,dx\) and \(v=\frac{x^2}{2}\). 2. Integration by parts gives \(F(x)=\frac{x^2}{2}\ln(x)-\frac{1}{2}\int x\,dx\) \(=\frac{x^2}{2}\ln(x)-\frac{x^2}{4}\). 3. Evaluate: \(F(e)-F(1)=\frac{e^2}{4}-\left(-\frac{1}{4}\right)=\frac{e^2+1}{4}\).

One antiderivative is \(F(x)=\frac{x^2}{2}\ln(x)-\frac{x^2}{4}\). The integral equals \(\frac{e^2+1}{4}\).

- Use the specified integration-by-parts choices before applying a trigonometric identity. - Evaluate the product term at both endpoints. - Replace \(\sin^2x\) by \(1-\cos^2x\) only after integration by parts has produced it.

1. With \(u=\cos x\) and \(\text{d}v=\cos x\,\text{d}x\), we have \(\text{d}u=-\sin x\,\text{d}x\) and \(v=\sin x\). 2. Integration by parts gives \(I=[\sin x\cos x]_0^\pi+\int_0^\pi\sin^2x\,\text{d}x\). 3. The boundary term is \(0\), and \(\sin^2x=1-\cos^2x\), so \(I=\pi-I\). 4. Hence \(2I=\pi\) and \(I=\frac{\pi}{2}\).

Integration by parts gives \(I=[\sin x\cos x]_0^\pi+\int_0^\pi\sin^2x\,\text{d}x=\pi-I\), so \(I=\frac{\pi}{2}\).

- Differentiate the polynomial factor so its degree decreases at each application of integration by parts. - Keep the common factor \(e^x\) when simplifying the antiderivative. - Determine the sign of \(x(x-2)e^x\) on the interval before converting signed accumulation to geometric area.

1. Apply integration by parts to \(\int(x^2-2x)e^x\,\text{d}x\) with \(u=x^2-2x\) and \(\text{d}v=e^x\,\text{d}x\). Then \(\text{d}u=(2x-2)\,\text{d}x\) and \(v=e^x\), so \(\int(x^2-2x)e^x\,\text{d}x=(x^2-2x)e^x-\int(2x-2)e^x\,\text{d}x\). 2. Apply integration by parts again to the remaining integral. An antiderivative is \(F(x)=(x^2-4x+4)e^x\). 3. Therefore, \(\int_0^2 f(x)\,\text{d}x=F(2)-F(0)=0-4=-4\). 4. On \([0,2]\), \(x(x-2)\le0\) and \(e^x>0\), so \(f(x)\le0\). The signed integral is negative, while geometric area is positive. Thus, the area is \(|-4|=4\) square units.

1) One antiderivative is \(F(x)=(x^2-4x+4)e^x\). 2) \(-4\) 3) \(4\) square units; the function is nonpositive on \([0,2]\), so the geometric area is the absolute value of the signed integral.

- For integration by parts, choose one factor whose derivative becomes simpler and let the other factor be \(1\). - Read the two marked endpoint labels and locate the x-intercept of the graph. - Split the geometric area where the graph changes sign. - A negative signed integral below the x-axis contributes positive geometric area.

1. For integration by parts, choose \(u=\ln(x)\) and \(\text{d}v=\text{d}x\). Then \(\text{d}u=\frac{1}{x}\text{d}x\) and \(v=x\). 2. Thus \(\int\ln(x)\,\text{d}x=x\ln(x)-\int1\,\text{d}x=x\ln(x)-x+C\). 3. The marked endpoints are \(x=\frac{1}{e}\) and \(x=e\). The graph crosses the x-axis at \(x=1\), so split the geometric area there. 4. On \([\frac{1}{e},1]\), the signed integral is \([x\ln(x)-x]_{1/e}^{1}=-1+\frac{2}{e}\), so the geometric area is \(1-\frac{2}{e}\). 5. On \([1,e]\), the graph is above the x-axis and the area is \([x\ln(x)-x]_1^e=1\). 6. The total area is \(2-\frac{2}{e}\approx1.264\) square units.

a) Using \(u=\ln(x)\) and \(\text{d}v=\text{d}x\), \(\int\ln(x)\,\text{d}x=x\ln(x)-x+C\). b) The area is \(2-\frac{2}{e}\approx1.264\) square units.

- View the single function as one factor multiplied by a constant function. - Differentiate the factor whose derivative becomes a rational expression. - After applying integration by parts, look for a denominator-derivative relationship.

1. Let \(u=\arctan(x)\) and \(\text{d}v=\text{d}x\), so \(\text{d}u=\frac{1}{1+x^2}\,\text{d}x\) and \(v=x\). 2. Integration by parts gives \(x\arctan(x)-\int\frac{x}{1+x^2}\,\text{d}x\). 3. The remaining integral is \(\frac{1}{2}\ln(1+x^2)\). 4. The antiderivative is \(x\arctan(x)-\frac{1}{2}\ln(1+x^2)+C\).

\(x\arctan(x)-\frac{1}{2}\ln(1+x^2)+C\)

- View the single inverse-trigonometric function as a product with \(1\). - Differentiate \(\arcsin(x)\) and integrate the constant factor. - In the remaining radical integral, compare the numerator with the derivative of \(1-x^2\).

1. Let \(u=\arcsin(x)\) and \(\text{d}v=\text{d}x\). Then \(\text{d}u=\frac{1}{\sqrt{1-x^2}}\,\text{d}x\) and \(v=x\). 2. Integration by parts gives \(x\arcsin(x)-\int\frac{x}{\sqrt{1-x^2}}\,\text{d}x\). 3. For the remaining integral, let \(w=1-x^2\). Then \(\int\frac{x}{\sqrt{1-x^2}}\,\text{d}x=-\sqrt{1-x^2}\). 4. Therefore, an antiderivative is \(x\arcsin(x)+\sqrt{1-x^2}+C\).

\(x\arcsin(x)+\sqrt{1-x^2}+C\)

- In the first application, differentiate \(x^2\) and integrate \(\cos(x)\). - The remaining integral still contains a product, so apply integration by parts again. - Differentiate the final expression to check cancellation of the extra terms.

1. Let \(u=x^2\) and \(\text{d}v=\cos(x)\,\text{d}x\). Then \(\text{d}u=2x\,\text{d}x\) and \(v=\sin(x)\). 2. The first application gives \(x^2\sin(x)-2\int x\sin(x)\,\text{d}x\). 3. For the remaining integral, let \(u=x\) and \(\text{d}v=\sin(x)\,\text{d}x\). Then \(\int x\sin(x)\,\text{d}x=-x\cos(x)+\sin(x)\). 4. Substitution gives \(x^2\sin(x)+2x\cos(x)-2\sin(x)+C\).

\(x^2\sin(x)+2x\cos(x)-2\sin(x)+C\)

- Treat the logarithm as one factor and pair it with \(1\). - After integration by parts, rewrite \(x/(x+1)\) as a constant minus a reciprocal term. - Evaluate the logarithmic boundary values exactly.

1. Let \(u=\ln(x+1)\) and \(\text{d}v=\text{d}x\). Then \(\text{d}u=\frac{1}{x+1}\,\text{d}x\) and \(v=x\). 2. Integration by parts gives \(\int_0^1\ln(x+1)\,\text{d}x=[x\ln(x+1)]_0^1-\int_0^1\frac{x}{x+1}\,\text{d}x\). 3. Rewrite \(\frac{x}{x+1}=1-\frac{1}{x+1}\). Therefore, \(\int_0^1\frac{x}{x+1}\,\text{d}x=[x-\ln(x+1)]_0^1=1-\ln2\). 4. The boundary term is \(\ln2\), so the integral equals \(\ln2-(1-\ln2)=2\ln2-1\).

\(2\ln2-1\)

- Differentiate the polynomial factor so its degree decreases. - Apply integration by parts again to the remaining linear–exponential product. - Factor out \(e^x\) when simplifying the result.

1. Let \(u=x^2\) and \(\text{d}v=e^x\,\text{d}x\). Then \(\text{d}u=2x\,\text{d}x\) and \(v=e^x\). 2. The first application gives \(x^2e^x-2\int xe^x\,\text{d}x\). 3. Apply integration by parts again to \(\int xe^x\,\text{d}x\): \(\int xe^x\,\text{d}x=xe^x-e^x\). 4. Substitute and simplify: \(x^2e^x-2(xe^x-e^x)=e^x(x^2-2x+2)+C\).

\(e^x(x^2-2x+2)+C\)

- Differentiate the inverse-tangent factor and integrate \(x\). - Rewrite the resulting rational expression by comparing its numerator with its denominator. - Combine the two inverse-tangent terms after integration.

1. Let \(u=\arctan(x)\) and \(\text{d}v=x\,\text{d}x\). Then \(\text{d}u=\frac{1}{1+x^2}\,\text{d}x\) and \(v=\frac{x^2}{2}\). 2. Integration by parts gives \(\frac{x^2}{2}\arctan(x)-\frac{1}{2}\int\frac{x^2}{1+x^2}\,\text{d}x\). 3. Rewrite \(\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}\). 4. Therefore, the antiderivative is \(\frac{x^2}{2}\arctan(x)-\frac{x}{2}+\frac{1}{2}\arctan(x)+C\), or \(\frac{x^2+1}{2}\arctan(x)-\frac{x}{2}+C\).

\(\frac{x^2+1}{2}\arctan(x)-\frac{x}{2}+C\)

- Differentiate the logarithmic factor and integrate \(x\). - After integration by parts, divide \(x^2\) by \(x+1\) before integrating. - Keep the boundary term separate until the logarithmic terms cancel.

1. Let \(u=\ln(x+1)\) and \(\text{d}v=x\,\text{d}x\). Then \(\text{d}u=\frac{1}{x+1}\,\text{d}x\) and \(v=\frac{x^2}{2}\). 2. Integration by parts gives \(\int_0^1x\ln(x+1)\,\text{d}x=\left[\frac{x^2}{2}\ln(x+1)\right]_0^1-\frac{1}{2}\int_0^1\frac{x^2}{x+1}\,\text{d}x\). 3. Divide: \(\frac{x^2}{x+1}=x-1+\frac{1}{x+1}\). Thus, \(\int_0^1\frac{x^2}{x+1}\,\text{d}x=\left[\frac{x^2}{2}-x+\ln(x+1)\right]_0^1=\ln2-\frac{1}{2}\). 4. Therefore, \(\frac{1}{2}\ln2-\frac{1}{2}\left(\ln2-\frac{1}{2}\right)=\frac{1}{4}\).

\(\frac{1}{4}\)

- Choose the factor whose derivative becomes simpler. - Keep the endpoint contribution separate from the remaining integral. - Evaluate each exact term before combining them.

1. Differentiate \(\ln(x)\) and integrate \(x^{-2}\). 2. Integration by parts gives \(\left[-\frac{\ln(x)}{x}\right]_1^e+\int_1^e\frac{1}{x^2}\,\text{d}x\). 3. The boundary term is \(-\frac{1}{e}\), and the remaining integral is \(1-\frac{1}{e}\). 4. The exact value is \(1-\frac{2}{e}\).

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

- Which factor becomes simpler each time it is differentiated? - After the first transformation, identify the remaining integral as a simpler version of the same pattern. - Check that the logarithmic terms cancel correctly when you differentiate.

1. Differentiate \([\ln(x)]^2\) and integrate \(x\). 2. Integration by parts gives \(\frac{x^2}{2}[\ln(x)]^2-\int x\ln(x)\,\text{d}x\). 3. A second application gives \(\int x\ln(x)\,\text{d}x=\frac{x^2}{2}\ln(x)-\frac{x^2}{4}\). 4. Therefore the antiderivative is \(\frac{x^2}{2}[\ln(x)]^2-\frac{x^2}{2}\ln(x)+\frac{x^2}{4}+C\).

\(\frac{x^2}{2}[\ln(x)]^2-\frac{x^2}{2}\ln(x)+\frac{x^2}{4}+C\)

- Treat the variable multiplying the second derivative as one factor. - The remaining integral can be determined from endpoint values. - Pay attention to the lower endpoint in the boundary product.

1. Integration by parts gives \(\int_0^1x f''(x)\,\text{d}x=[xf'(x)]_0^1-\int_0^1f'(x)\,\text{d}x\). 2. The boundary product is \(f'(1)=4\). 3. The remaining integral is \(f(1)-f(0)=5-2=3\). 4. The requested value is \(4-3=1\).

\(1\)

- Transform the integral containing the derivative into a boundary term and an integral without the derivative. - Evaluate the boundary expression before solving for the unknown integral. - Use the given integral value as part of an equation.

1. Integration by parts gives \(\int_0^2xg'(x)\,\text{d}x=[xg(x)]_0^2-\int_0^2g(x)\,\text{d}x\). 2. The boundary product is \(2g(2)-0g(0)=10\). 3. Substitute the given value: \(7=10-\int_0^2g(x)\,\text{d}x\). 4. Therefore \(\int_0^2g(x)\,\text{d}x=3\).

\(3\)

- Treat the inverse-tangent function as a product with \(1\). - After integration by parts, compare the remaining numerator with the derivative of \(1+x^2\). - Use \(\arctan(1)=\pi/4\) and \(\arctan(0)=0\).

1. Let \(u=\arctan(x)\) and \(\text{d}v=\text{d}x\). Then \(\text{d}u=\frac{1}{1+x^2}\,\text{d}x\) and \(v=x\). 2. Integration by parts gives \(\int_0^1\arctan(x)\,\text{d}x=[x\arctan(x)]_0^1-\int_0^1\frac{x}{1+x^2}\,\text{d}x\). 3. The boundary term is \(\frac{\pi}{4}\), and \(\int_0^1\frac{x}{1+x^2}\,\text{d}x=\frac{1}{2}\ln2\). 4. Therefore, the exact value is \(\frac{\pi}{4}-\frac{1}{2}\ln2\).

\(\frac{\pi}{4}-\frac{1}{2}\ln2\)

- Treat the logarithm as one factor multiplied by \(1\). - After integration by parts, rewrite the rational expression as a constant minus \(1/(1+x^2)\). - Differentiate the final expression to check the cancellation of the rational terms.

1. Let \(u=\ln(1+x^2)\) and \(\text{d}v=\text{d}x\). Then \(\text{d}u=\frac{2x}{1+x^2}\,\text{d}x\) and \(v=x\). 2. Integration by parts gives \(x\ln(1+x^2)-2\int\frac{x^2}{1+x^2}\,\text{d}x\). 3. Rewrite \(\frac{x^2}{1+x^2}=1-\frac{1}{1+x^2}\). 4. Therefore, the antiderivative is \(x\ln(1+x^2)-2x+2\arctan(x)+C\).

\(x\ln(1+x^2)-2x+2\arctan(x)+C\)

- Separate the term containing the unknown constant. - One of the definite integrals requires integration by parts. - After evaluating both integrals, solve the resulting linear equation.

1. Split the integral as \(\int_0^1xe^x\,\text{d}x-k\int_0^1e^x\,\text{d}x\). 2. Integration by parts gives \(\int_0^1xe^x\,\text{d}x=[xe^x]_0^1-\int_0^1e^x\,\text{d}x=1\). 3. Also, \(\int_0^1e^x\,\text{d}x=e-1\). 4. The condition becomes \(1-k(e-1)=0\), so \(k=\frac{1}{e-1}\).

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

- Apply integration by parts more than once. - Differentiating a power of \(\ln(t)\) lowers that power. - Track the signs when substituting the second integration-by-parts result. - Evaluate all logarithmic terms at the lower limit \(1\).

1. Use integration by parts with \(u=[\ln(t)]^2\) and \(dv=dt\). Then \(du=\frac{2\ln(t)}{t}\,dt\) and \(v=t\). 2. Therefore, \(\int[\ln(t)]^2\,dt=t[\ln(t)]^2-2\int\ln(t)\,dt\). 3. Apply integration by parts again to \(\int\ln(t)\,dt\), obtaining \(t\ln(t)-t\). 4. An antiderivative is \(t[\ln(t)]^2-2t\ln(t)+2t\). 5. Evaluate from \(1\) to \(x\). Since \(\ln(1)=0\), \(I_1(x)=x[\ln(x)]^2-2x\ln(x)+2x-2\).

\(I_1(x)=x[\ln(x)]^2-2x\ln(x)+2x-2\)

- Apply integration by parts more than once. - Track the signs when differentiating cosine. - When the original integral returns, move it to the same side of the equation. - Use the sine and cosine values at \(0\) and \(\pi\).

1. Let \(I=\int_0^\pi e^x\sin(x)\,dx\). Use integration by parts with \(u=\sin(x)\) and \(dv=e^x\,dx\): \(I=[e^x\sin(x)]_0^\pi-\int_0^\pi e^x\cos(x)\,dx\). The boundary term is \(0\), so \(I=-\int_0^\pi e^x\cos(x)\,dx\). 2. Apply integration by parts to the remaining integral with \(u=\cos(x)\) and \(dv=e^x\,dx\): \(\int_0^\pi e^x\cos(x)\,dx=[e^x\cos(x)]_0^\pi+I\). 3. Substitute into the first equation: \(I=-[e^x\cos(x)]_0^\pi-I\). Thus \(2I=-[e^x\cos(x)]_0^\pi\). 4. Evaluate the boundary term: \(-[e^\pi\cos(\pi)-e^0\cos(0)]=e^\pi+1\). Therefore, \(I=\frac{e^\pi+1}{2}\).

\(\frac{e^\pi+1}{2}\)

- Insert a factor of \(1\) to create a product. - Apply integration by parts again to the remaining logarithmic integral. - Track the minus sign carefully. - Recall the derivative of \(\ln(x)\).

1. Let \(u=(\ln(x))^2\) and \(dv=dx\). Then \(du=\frac{2\ln(x)}{x}\,dx\) and \(v=x\). 2. Thus, \(\int(\ln(x))^2\,dx=x(\ln(x))^2-2\int\ln(x)\,dx\). 3. Apply integration by parts again to \(\int\ln(x)\,dx\): \(\int\ln(x)\,dx=x\ln(x)-x\). 4. Therefore, an antiderivative is \(F(x)=x(\ln(x))^2-2x\ln(x)+2x\). 5. Evaluate: \([F(x)]_1^e=e-2\).

\(e-2\)

- Differentiate the polynomial factor so its degree decreases. - Apply integration by parts twice. - Track the signs when integrating sine and cosine. - Use the exact trigonometric values at \(0\) and \(\pi\).

1. Let \(u=x^2-2x\) and \(dv=\sin(x)\,dx\). Then \(du=(2x-2)\,dx\) and \(v=-\cos(x)\). 2. Therefore, \(\int(x^2-2x)\sin(x)\,dx=-(x^2-2x)\cos(x)+\int(2x-2)\cos(x)\,dx\). 3. Apply integration by parts again with \(u=2x-2\) and \(dv=\cos(x)\,dx\): \(\int(2x-2)\cos(x)\,dx=(2x-2)\sin(x)+2\cos(x)\). 4. An antiderivative is \(F(x)=-(x^2-2x)\cos(x)+(2x-2)\sin(x)+2\cos(x)\). 5. Evaluate: \(F(\pi)=\pi^2-2\pi-2\) and \(F(0)=2\), so the integral is \(\pi^2-2\pi-4\).

\(\pi^2-2\pi-4\approx -0.414\)

- Differentiate the exponential factor and integrate the trigonometric factor twice. - Treat the recurring integral as an unknown quantity. - Solve the resulting equation for that integral. - Use the exact sine and cosine values at the bounds.

1. Let \(I=\int e^{2x}\cos(x)\,dx\). With \(u=e^{2x}\) and \(dv=\cos(x)\,dx\), \(I=e^{2x}\sin(x)-2\int e^{2x}\sin(x)\,dx\). 2. Apply integration by parts to the remaining integral: \(\int e^{2x}\sin(x)\,dx=-e^{2x}\cos(x)+2I\). 3. Substitute this expression: \(I=e^{2x}\sin(x)+2e^{2x}\cos(x)-4I\). Thus, \(I=\frac{1}{5}e^{2x}(\sin(x)+2\cos(x))\). 4. Evaluate from \(0\) to \(\frac{\pi}{2}\): \(\frac{1}{5}e^\pi-\frac{2}{5}=\frac{e^\pi-2}{5}\).

\(\frac{e^\pi-2}{5}\approx 4.228\)

- Differentiate the entire product before comparing it with the given expression. - Organize the result by powers of the variable. - The coefficients of matching powers must agree.

1. Differentiate the proposed product to obtain \(e^{2x}[2ax^2+(2b+2a)x+(2c+b)]\). 2. Match coefficients with \(e^{2x}(3x^2-4x+5)\): \(2a=3\), \(2b+2a=-4\), and \(2c+b=5\). 3. Solving gives \(a=\frac{3}{2}\), \(b=-\frac{7}{2}\), and \(c=\frac{17}{4}\). 4. Therefore the integral is \(e^{2x}\left(\frac{3}{2}x^2-\frac{7}{2}x+\frac{17}{4}\right)+C\).

\(a=\frac{3}{2}\), \(b=-\frac{7}{2}\), and \(c=\frac{17}{4}\) \(\int e^{2x}(3x^2-4x+5)\,\text{d}x=e^{2x}\left(\frac{3}{2}x^2-\frac{7}{2}x+\frac{17}{4}\right)+C\)

- Choose the logarithm as the factor to differentiate. - After the first transformation, rewrite the remaining rational expression as simpler fractions. - Use the stated domain when simplifying logarithmic absolute values.

1. Differentiate \(\ln(x)\) and integrate \((1+x)^{-2}\). 2. Integration by parts gives \(-\frac{\ln(x)}{1+x}+\int\frac{1}{x(1+x)}\,\text{d}x\). 3. Decompose \(\frac{1}{x(1+x)}=\frac{1}{x}-\frac{1}{1+x}\). 4. The antiderivative is \(-\frac{\ln(x)}{1+x}+\ln(x)-\ln(1+x)+C\).

\(-\frac{\ln(x)}{1+x}+\ln(x)-\ln(1+x)+C\)