Free math worksheets

- Compare numerator and denominator degrees before choosing any other method. - For a quadratic denominator, ask whether rewriting it around its vertex reveals a standard form. - Also check whether the numerator already relates directly to the denominator’s derivative.

1. a) The numerator's degree is at least the denominator's degree, so use polynomial long division. 2. b) The quadratic denominator is irreducible in its given form and becomes \((x+3)^2+4\), so complete the square. 3. c) The numerator is the derivative of the denominator up to a constant factor, so neither listed rearrangement is needed. 4. d) The numerator matches half the derivative of the denominator, so neither rearrangement is needed before integrating.

a) Polynomial long division b) Completing the square c) Neither d) Neither

- Rewrite the quadratic as a squared binomial plus a positive constant. - Identify the scale of the constant term after the square is completed. - Account for that scale both inside and outside the inverse trigonometric expression.

1. Rewrite the denominator: \(x^2+6x+13=(x+3)^2+4\). 2. Use \(u=x+3\) and the standard form \(\int\frac{1}{u^2+a^2}\,\text{d}u=\frac{1}{a}\arctan\left(\frac{u}{a}\right)+C\) with \(a=2\). 3. The antiderivative is \(\frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C\).

\(\frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C\)

- Use division because the numerator and denominator have comparable degrees. - Check that the logarithm’s argument stays positive on the interval. - Evaluate the polynomial and logarithmic parts at both bounds.

1. Divide: \(\frac{x^2+2x+4}{x+1}=x+1+\frac{3}{x+1}\). 2. An antiderivative on \([0, 2]\) is \(\frac{x^2}{2}+x+3\ln(x+1)\). 3. Evaluating at the bounds gives \(\left(2+2+3\ln3\right)-0=4+3\ln3\).

\(4+3\ln3\)

- Rewrite the denominator as a shifted square plus one. - Find the full antiderivative family before applying the initial condition. - Use the condition to determine the additive constant exactly.

1. Complete the square: \(x^2+4x+5=(x+2)^2+1\). 2. The antiderivative family is \(F(x)=\arctan(x+2)+C\). 3. The condition gives \(0=\arctan(2)+C\), so \(C=-\arctan(2)\). 4. Thus \(F(x)=\arctan(x+2)-\arctan(2)\).

\(F(x)=\arctan(x+2)-\arctan(2)\)

- Rewrite the rational function in quotient-remainder form before inspecting the choices. - Compare the polynomial part of each candidate with the quotient. - Differentiate the logarithmic coefficient to check the remainder term.

1. Divide: \(\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}\). 2. Integrating gives \(\frac{x^2}{2}-x+2\ln|x+1|+C\). 3. Candidate A has derivative \(x-1+\frac{2}{x+1}\), so A is correct.

A. \(\frac{x^2}{2}-x+2\ln|x+1|\)

- Add and subtract the square of half the linear coefficient. - Match the completed denominator to \(u^2+a^2\). - Account for the scale \(a=2\) in the inverse-tangent antiderivative.

1. Complete the square: \(x^2-6x+13=(x-3)^2+4\). 2. Rewrite the denominator as \((x-3)^2+2^2\). Using the inverse-tangent pattern, \(\int\frac{1}{(x-3)^2+2^2}\,dx=\frac12\arctan\left(\frac{x-3}{2}\right)+C\).

\(\frac12\arctan\left(\frac{x-3}{2}\right)+C\)

- The numerator has degree at least as large as the denominator, so divide first. - Check the quotient and remainder by multiplying back. - Integrate the remainder term using the logarithmic pattern.

1. Divide \(x^2+1\) by \(x-1\): \(\frac{x^2+1}{x-1}=x+1+\frac{2}{x-1}\). 2. Integrate the polynomial and reciprocal terms separately: \(\int\frac{x^2+1}{x-1}\,dx=\frac{x^2}{2}+x+2\ln|x-1|+C\).

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

- Verify a proposed division result by recombining quotient, divisor, and remainder. - Include a zero coefficient for the missing squared term when organizing the division. - After correcting the remainder, separate its derivative-like and constant parts.

1. Multiplying the claimed quotient by the divisor and adding the claimed remainder does not reproduce the numerator. 2. Correct division gives \(\frac{2x^3+x+4}{x^2+1}=2x+\frac{-x+4}{x^2+1}\). 3. Integrate: \(\int2x\,\text{d}x=x^2\), \(\int\frac{-x}{x^2+1}\,\text{d}x=-\frac{1}{2}\ln(x^2+1)\), and \(\int\frac{4}{x^2+1}\,\text{d}x=4\arctan(x)\). 4. The result is \(x^2-\frac{1}{2}\ln(x^2+1)+4\arctan(x)+C\).

Correct decomposition: \(2x+\frac{-x+4}{x^2+1}\) Antiderivative: \(x^2-\frac{1}{2}\ln(x^2+1)+4\arctan(x)+C\)

- Compare the numerator with the derivative of the quadratic denominator. - Separate any leftover constant before integrating. - Rewrite the quadratic so the remaining proper fraction matches a familiar shifted form.

1. The denominator derivative is \(2x+4\), so write \(2x+5=(2x+4)+1\). 2. Complete the square: \(x^2+4x+8=(x+2)^2+4\). 3. The derivative-matching part integrates to \(\ln(x^2+4x+8)\). 4. The remaining part integrates to \(\frac{1}{2}\arctan\left(\frac{x+2}{2}\right)\). 5. The antiderivative is \(\ln(x^2+4x+8)+\frac{1}{2}\arctan\left(\frac{x+2}{2}\right)+C\).

\(\ln(x^2+4x+8)+\frac{1}{2}\arctan\left(\frac{x+2}{2}\right)+C\)

- Rewrite the rational function as a polynomial and a proper fraction. - Use the stated domain to simplify the logarithm’s absolute value. - Apply the point condition only after finding the full antiderivative family.

1. Divide: \(\frac{x^2-1}{x-2}=x+2+\frac{3}{x-2}\). 2. On \(x>2\), \(F(x)=\frac{x^2}{2}+2x+3\ln(x-2)+C\). 3. Use \(F(3)=5\): \(\frac{9}{2}+6+3\ln1+C=5\). 4. Thus \(C=-\frac{11}{2}\), so \(F(x)=\frac{x^2}{2}+2x+3\ln(x-2)-\frac{11}{2}\).

\(F(x)=\frac{x^2}{2}+2x+3\ln(x-2)-\frac{11}{2}\), for \(x>2\)

- Use the divisor’s zero to determine the division remainder efficiently. - Require that remainder to vanish before integrating. - After the parameter is fixed, verify that the quotient is a polynomial.

1. The remainder on division by \(x+2\) is the numerator evaluated at \(x=-2\): \(-8-2k+6=-2-2k\). 2. Set the remainder equal to zero: \(-2-2k=0\), so \(k=-1\). 3. For \(k=-1\), division gives \(\frac{x^3-x+6}{x+2}=x^2-2x+3\). 4. The integral is \(\frac{x^3}{3}-x^2+3x+C\).

\(k=-1\), and the integral is \(\frac{x^3}{3}-x^2+3x+C\).

- Use the relationship among dividend, divisor, quotient, and remainder. - Keep the given quotient-remainder form for the integration step. - Integrate the polynomial and proper-fraction terms separately.

1. Numerator equals divisor times quotient plus remainder: \((x+3)(2x-1)+5=2x^2+5x+2\). 2. The rational function is already decomposed as \(2x-1+\frac{5}{x+3}\). 3. Its antiderivative is \(x^2-x+5\ln|x+3|+C\).

a) \(2x^2+5x+2\) b) \(x^2-x+5\ln|x+3|+C\)

- Compare each numerator with the derivative of the common denominator. - For the constant numerator, rewrite the quadratic around its vertex. - Do not perform an algebraic rearrangement when a direct derivative relationship already resolves the integral.

1. a) The numerator is the derivative of the denominator, so the result is \(\ln(x^2-6x+13)+C\). 2. b) Complete the square: \(x^2-6x+13=(x-3)^2+4\). 3. Therefore, b) is \(\frac{1}{2}\arctan\left(\frac{x-3}{2}\right)+C\). 4. Completing the square is unnecessary in a) because the derivative match determines the antiderivative directly.

a) \(\ln(x^2-6x+13)+C\) b) \(\frac{1}{2}\arctan\left(\frac{x-3}{2}\right)+C\)

- Rewrite the quadratic denominator as a squared horizontal shift plus a positive constant. - Compare the intervals \([2,4]\) and \([4,6]\) after the substitution \(u=x-4\). - Use the standard antiderivative for \(1/(u^2+a^2)\) with \(a=2\).

1. Complete the square: \(x^2-8x+20=(x-4)^2+4\). 2. The integrand depends on \((x-4)^2\), so its graph is symmetric about \(x=4\). Therefore, \(\int_2^6\frac{1}{(x-4)^2+4}\,\text{d}x=2\int_4^6\frac{1}{(x-4)^2+4}\,\text{d}x\). 3. An antiderivative is \(\frac{1}{2}\arctan\left(\frac{x-4}{2}\right)\). 4. Thus, \(2\left[\frac{1}{2}\arctan\left(\frac{x-4}{2}\right)\right]_4^6=\arctan(1)-\arctan(0)=\frac{\pi}{4}\).

\(\frac{\pi}{4}\)

- Factor the negative sign from the quadratic terms before completing the square. - Rewrite the radicand as a positive constant square minus a shifted variable square. - Check whether the inner shifted expression has derivative one.

1. Rewrite the radicand: \(-x^2+4x+5=9-(x-2)^2\). 2. This matches the form \(a^2-u^2\) with \(a=3\) and \(u=x-2\). 3. The antiderivative is \(\arcsin\left(\frac{x-2}{3}\right)+C\).

\(\arcsin\left(\frac{x-2}{3}\right)+C\)

- Check the proposed result by differentiating it rather than repeating the integration. - Track the derivative of the linear expression inside the inverse tangent. - Adjust the outside coefficient so the derivative matches the original integrand exactly.

1. Differentiating the claim introduces a factor of \(2\) from \(2x+3\), so the derivative is twice the required integrand. 2. Equivalently, with \(u=2x+3\), \(\text{d}x=\frac{1}{2}\text{d}u\). 3. The correct antiderivative is \(\frac{1}{4}\arctan\left(\frac{2x+3}{2}\right)+C\).

The claimed antiderivative is too large by a factor of \(2\). The correct result is \(\frac{1}{4}\arctan\left(\frac{2x+3}{2}\right)+C\).

- Use the divisor’s root to express the remainder in terms of the parameter. - Set that expression equal to the prescribed remainder. - Keep the nonzero remainder as a proper fraction after division.

1. The remainder is the numerator evaluated at \(x=-1\): \(-1+2+k=1+k\). 2. Set \(1+k=3\), giving \(k=2\). 3. Division gives \(\frac{x^3+2x^2+2}{x+1}=x^2+x-1+\frac{3}{x+1}\). 4. The integral is \(\frac{x^3}{3}+\frac{x^2}{2}-x+3\ln|x+1|+C\).

\(k=2\), and the antiderivative is \(\frac{x^3}{3}+\frac{x^2}{2}-x+3\ln|x+1|+C\).

- Differentiate the polynomial and logarithmic parts separately. - Treat the derivative as the desired quotient-remainder form. - Recombine the polynomial and proper fraction over a common denominator.

1. Differentiate: \(F'(x)=x^2-2x+\frac{4}{x+2}\). 2. This is the quotient-remainder form. 3. Combine into one fraction: \(x^2-2x+\frac{4}{x+2}=\frac{(x^2-2x)(x+2)+4}{x+2}=\frac{x^3-4x+4}{x+2}\).

\(f(x)=\frac{x^3-4x+4}{x+2}\), with quotient-remainder form \(f(x)=x^2-2x+\frac{4}{x+2}\).

- Continue the division until the remainder’s degree is strictly less than the denominator’s. - Notice that the final remainder is related to the derivative of the quadratic denominator. - Evaluate the logarithmic term carefully at the lower bound.

1. Divide: \(\frac{x^3+x^2+1}{x^2+1}=x+1-\frac{x}{x^2+1}\). 2. An antiderivative is \(\frac{x^2}{2}+x-\frac{1}{2}\ln(x^2+1)\). 3. Evaluating from \(0\) to \(1\) gives \(\frac{1}{2}+1-\frac{1}{2}\ln2=\frac{3}{2}-\frac{1}{2}\ln2\).

\(\frac{3}{2}-\frac{1}{2}\ln2\)

- Complete the square while keeping the leading coefficient incorporated into the binomial. - Track the derivative of the resulting linear expression. - Include both the substitution scale and the quadratic scale in the outside coefficient.

1. Complete the square: \(9x^2-12x+8=(3x-2)^2+4\). 2. Let \(u=3x-2\), so \(\text{d}x=\frac{1}{3}\text{d}u\). 3. The integral becomes \(\frac{1}{3}\int\frac{1}{u^2+4}\,\text{d}u\). 4. The result is \(\frac{1}{6}\arctan\left(\frac{3x-2}{2}\right)+C\).

\(\frac{1}{6}\arctan\left(\frac{3x-2}{2}\right)+C\)

- Rewrite the quadratic under the radical as a constant minus a square. - Use the center, endpoints, and maximum height of the displayed graph to identify its geometric shape. - Check that the integration bounds cover the full horizontal diameter.

1. Complete the square: \(-x^2-2x+15=16-(x+1)^2\). 2. The graph \(y=\sqrt{16-(x+1)^2}\) is the upper semicircle of radius \(4\), centered at \((-1, 0)\). 3. The interval \([-5, 3]\) spans the entire semicircle. 4. The integral is its area: \(\frac{1}{2}\pi(4)^2=8\pi\).

\(8\pi\)

- Translate the location and value of the minimum into vertex form. - Expand the completed square only after identifying the quadratic. - Compare the numerator of the integral with the derivative of its denominator.

1. The vertex information gives \(q(x)=(x+2)^2+3\). 2. Expanding gives \(q(x)=x^2+4x+7\), so \(b=4\) and \(c=7\). 3. Its derivative is \(q'(x)=2x+4\). 4. Therefore \(\int\frac{q'(x)}{q(x)}\,\text{d}x=\ln(q(x))+C=\ln(x^2+4x+7)+C\).

\(b=4\), \(c=7\), and \(\int\frac{q'(x)}{q(x)}\,\text{d}x=\ln(x^2+4x+7)+C\).

- Write every polynomial degree in descending order, inserting zero coefficients where needed. - Continue division until the remainder degree is below the denominator degree. - Split the final linear remainder into derivative-related and constant pieces.

1. The dividend is \(x^4+0x^3+0x^2-3x+2\); the omitted \(0x^2\) term affects the subtraction after the first quotient term. 2. Correct division gives \(\frac{x^4-3x+2}{x^2+1}=x^2-1+\frac{-3x+3}{x^2+1}\). 3. Integrate the polynomial part: \(\frac{x^3}{3}-x\). 4. The remainder gives \(-\frac{3}{2}\ln(x^2+1)+3\arctan(x)\). 5. The result is \(\frac{x^3}{3}-x-\frac{3}{2}\ln(x^2+1)+3\arctan(x)+C\).

Correct decomposition: \(x^2-1+\frac{-3x+3}{x^2+1}\) Antiderivative: \(\frac{x^3}{3}-x-\frac{3}{2}\ln(x^2+1)+3\arctan(x)+C\)

- Divide first so each component has an easily recognized symmetry. - Separate the odd and even remainder terms. - Use the symmetric bounds to eliminate one contribution before evaluating the others.

1. Divide: \(\frac{x^2+2x+5}{x^2+1}=1+\frac{2x+4}{x^2+1}\). 2. The term \(\frac{2x}{x^2+1}\) is odd, so its integral over \([-1, 1]\) is \(0\). 3. The constant term contributes \(2\). 4. The remaining even term gives \(4\int_{-1}^{1}\frac{1}{x^2+1}\,\text{d}x=4\left(\frac{\pi}{2}\right)=2\pi\). 5. The integral is \(2+2\pi\).

\(2+2\pi\)

- Use polynomial division to separate a polynomial from a multiple of \(1/(x^2+1)\). - Evaluate the polynomial part over the symmetric interval before introducing the parameter equation. - Use \(\int_{-1}^{1}1/(x^2+1)\,\text{d}x=\pi/2\), then solve the resulting linear equation in \(k\).

1. Polynomial division gives \(\frac{x^4+k}{x^2+1}=x^2-1+\frac{k+1}{x^2+1}\). 2. Over \([-1,1]\), \(\int_{-1}^{1}(x^2-1)\,\text{d}x=\frac{2}{3}-2=-\frac{4}{3}\). 3. Also, \(\int_{-1}^{1}\frac{1}{x^2+1}\,\text{d}x=\frac{\pi}{2}\). Therefore, the given integral equals \(-\frac{4}{3}+(k+1)\frac{\pi}{2}\). 4. Set this equal to \(2\pi-\frac{4}{3}\). Then \((k+1)\frac{\pi}{2}=2\pi\), so \(k+1=4\) and \(k=3\).

\(k=3\)

- Rewrite the improper rational function as a polynomial plus a proper fraction. - Compare the remaining numerator with the derivative of the denominator. - For the verification, combine the derivative terms over a common denominator.

1. Divide: \(\frac{x^3-2x}{x^2+1}=x-\frac{3x}{x^2+1}\). 2. Integrate: \(\int x\,\text{d}x=\frac{x^2}{2}\), and \(\int\frac{-3x}{x^2+1}\,\text{d}x=-\frac{3}{2}\ln(x^2+1)\). 3. The antiderivative is \(\frac{x^2}{2}-\frac{3}{2}\ln(x^2+1)+C\). 4. Differentiating gives \(x-\frac{3x}{x^2+1}=\frac{x^3-2x}{x^2+1}\).

\(\frac{x^2}{2}-\frac{3}{2}\ln(x^2+1)+C\)

- Use the completed-square form to find the denominator’s minimum value. - Only after confirming positivity should you treat the integral as proper. - Evaluate the shifted inverse-tangent antiderivative at both endpoints.

1. Complete the square: \(x^2+6x+10=(x+3)^2+1\), which is positive for every real \(x\). Thus there are no singularities. 2. An antiderivative is \(\arctan(x+3)\). 3. Evaluating gives \(\arctan(8)-\arctan(-2)=\arctan(8)+\arctan(2)\).

There are no singularities on \([-5, 5]\), and the value is \(\arctan(8)+\arctan(2)\).

- Compare polynomial degrees in the first integrand before inspecting its remainder. - Rewrite the second quadratic around its vertex. - After division in part a), separate the linear remainder into two standard pieces.

1. a) Long division gives \(\frac{x^3+5x+1}{x^2+4}=x+\frac{x+1}{x^2+4}\). 2. Integrating gives \(\frac{x^2}{2}+\frac{1}{2}\ln(x^2+4)+\frac{1}{2}\arctan\left(\frac{x}{2}\right)+C\). 3. b) Complete the square: \(x^2+10x+29=(x+5)^2+4\). 4. Therefore, b) is \(\frac{1}{2}\arctan\left(\frac{x+5}{2}\right)+C\).

a) \(\frac{x^2}{2}+\frac{1}{2}\ln(x^2+4)+\frac{1}{2}\arctan\left(\frac{x}{2}\right)+C\) b) \(\frac{1}{2}\arctan\left(\frac{x+5}{2}\right)+C\)

- Divide first so the remaining numerator is linear. - A logarithm alone is produced when the remainder numerator is a constant multiple of the denominator’s derivative. - Match the x-coefficient first, then use the constant term to determine \(n\).

1. Divide: \(\frac{x^2+8x+n}{x^2+6x+13}=1+\frac{2x+n-13}{x^2+6x+13}\). 2. An inverse-trigonometric term is unnecessary when the remainder numerator is a constant multiple of the denominator derivative \(2x+6\). 3. The coefficient of \(x\) in the remainder is already \(2\), so the required multiple is \(1\). Therefore, \(n-13=6\), giving \(n=19\). 4. For \(n=19\), \(\frac{x^2+8x+19}{x^2+6x+13}=1+\frac{2x+6}{x^2+6x+13}\). Thus, the antiderivative is \(x+\ln(x^2+6x+13)+C\).

\(n=19\) \(\int\frac{x^2+8x+19}{x^2+6x+13}\,\text{d}x=x+\ln(x^2+6x+13)+C\)

- Split the numerator into a multiple of the denominator’s derivative and a constant remainder. - Complete the square to identify the scale of the inverse-tangent term. - Compare the two resulting coefficients only after both scale factors are included.

1. Write \(x+k=\frac{1}{2}(2x+2)+(k-1)\). 2. Since \(x^2+2x+5=(x+1)^2+4\), an antiderivative is \(\frac{1}{2}\ln(x^2+2x+5)+\frac{k-1}{2}\arctan\left(\frac{x+1}{2}\right)+C\). 3. Equal coefficients require \(\frac{k-1}{2}=\frac{1}{2}\), so \(k=2\). 4. The resulting antiderivative is \(\frac{1}{2}\ln(x^2+2x+5)+\frac{1}{2}\arctan\left(\frac{x+1}{2}\right)+C\).

\(k=2\) \(\frac{1}{2}\ln(x^2+2x+5)+\frac{1}{2}\arctan\left(\frac{x+1}{2}\right)+C\)

- Shift the variable to the center of the completed square. - Separate the transformed numerator into odd and even contributions. - Only the even contribution survives on symmetric bounds.

1. Let \(u=x+2\). Then the bounds become \(-2\) and \(2\), the denominator becomes \(u^2+6\), and the numerator becomes \(3u+1\). 2. The term \(\frac{3u}{u^2+6}\) is odd, so its integral over \([-2, 2]\) is \(0\). 3. The remaining integral is \(\int_{-2}^{2}\frac{1}{u^2+6}\,\text{d}u\). 4. Its value is \(\frac{2}{\sqrt{6}}\arctan\left(\frac{2}{\sqrt{6}}\right)\).

\(\frac{2}{\sqrt{6}}\arctan\left(\frac{2}{\sqrt{6}}\right)\)

- Expand the proposed completed-square form and compare coefficients. - The constant-term condition has two real solutions for the shift. - Each completed square has the same positive constant but a different center.

1. Expanding \((x-h)^2+4\) gives \(x^2-2hx+h^2+4\). 2. Matching coefficients gives \(b=-2h\) and \(h^2+4=13\). 3. Thus \(h=3\) or \(h=-3\), producing \((b,h)=(-6,3)\) and \((6, -3)\). 4. For \((b,h)=(-6,3)\), the integral is \(\frac{1}{2}\arctan\left(\frac{x-3}{2}\right)+C\). 5. For \((b,h)=(6,-3)\), the integral is \(\frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C\).

\((b,h)=(-6,3)\): \(\frac{1}{2}\arctan\left(\frac{x-3}{2}\right)+C\) \((b,h)=(6,-3)\): \(\frac{1}{2}\arctan\left(\frac{x+3}{2}\right)+C\)

- Differentiate the proposed antiderivative rather than integrating from scratch. - Write both derivative terms over the common quadratic denominator. - Match coefficients in the resulting linear numerator.

1. Differentiate the logarithmic term: \(A\frac{2x+4}{x^2+4x+8}\). 2. Differentiate the inverse-tangent term: \(B\frac{2}{x^2+4x+8}\). 3. Match numerators: \(A(2x+4)+2B=4x+10\). 4. Comparing x-coefficients gives \(2A=4\), so \(A=2\). Comparing constants gives \(8+2B=10\), so \(B=1\).

\(A=2\) and \(B=1\)

- Start with division because the numerator and denominator have the same degree. - Express the linear remainder using the denominator’s derivative plus a constant. - Complete the square only for the remaining constant-over-quadratic term.

1. Divide: \(\frac{3x^2+x+7}{x^2+2x+5}=3+\frac{-5x-8}{x^2+2x+5}\). 2. Write \(-5x-8=-\frac{5}{2}(2x+2)-3\). 3. Complete the square: \(x^2+2x+5=(x+1)^2+4\). 4. Integrate to get \(3x-\frac{5}{2}\ln(x^2+2x+5)-\frac{3}{2}\arctan\left(\frac{x+1}{2}\right)+C\).

\(3x-\frac{5}{2}\ln(x^2+2x+5)-\frac{3}{2}\arctan\left(\frac{x+1}{2}\right)+C\)

- Use division to obtain an accumulation formula in terms of the upper bound. - Look for a simple positive value that reproduces both the algebraic and logarithmic parts of the target. - Use the sign of the integrand to justify that no second positive solution exists.

1. Divide: \(\frac{x^2+2x+2}{x+1}=x+1+\frac{1}{x+1}\). 2. For \(b>0\), the integral equals \(\frac{b^2}{2}+b+\ln(b+1)\). 3. Substituting \(b=1\) gives \(\frac{1}{2}+1+\ln2=\frac{3}{2}+\ln2\), so \(b=1\). 4. The integrand is positive for \(x\ge0\), so the accumulated integral is strictly increasing in \(b\). Therefore, the solution is unique.

\(b=1\)