Free math worksheets

- Differentiate each logarithmic term separately. - Combine the resulting simple fractions using one common denominator. - Check the numerator after distributing both coefficients.

1. Differentiate the logarithms: \(F'(x)=\frac{3}{x-2}-\frac{2}{x+1}\). 2. Use the common denominator \((x-2)(x+1)\). 3. The numerator is \(3(x+1)-2(x-2)=x+7\). 4. Therefore \(F'(x)=\frac{x+7}{(x-2)(x+1)}\), and its indefinite integral is \(3\ln|x-2|-2\ln|x+1|+C\).

\(F'(x)=\frac{x+7}{(x-2)(x+1)}\) \(\int\frac{x+7}{(x-2)(x+1)}\,\text{d}x=3\ln|x-2|-2\ln|x+1|+C\)

- The denominator must be written as distinct linear factors first. - Use each root to isolate one numerator constant. - Recombine the fractions before integrating if a coefficient sign seems uncertain.

1. Factor \(x^2-x-6=(x-3)(x+2)\). 2. Write \(\frac{3x+2}{(x-3)(x+2)}=\frac{A}{x-3}+\frac{B}{x+2}\). 3. Substituting the roots gives \(A=\frac{11}{5}\) and \(B=\frac{4}{5}\). 4. The antiderivative is \(\frac{11}{5}\ln|x-3|+\frac{4}{5}\ln|x+2|+C\).

\(\frac{11}{5}\ln|x-3|+\frac{4}{5}\ln|x+2|+C\)

- A repeated linear factor requires one term for each power of that factor. - Recombine the two proposed terms and match the numerator coefficients. - Integrate the first term logarithmically and the second as a negative power.

1. Use the repeated-factor form \(\frac{3x+5}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}\). 2. Recombining gives \(3x+5=A(x-1)+B\), so \(A=3\) and \(B=8\). 3. Integrate term by term: \(\int\left[\frac{3}{x-1}+\frac{8}{(x-1)^2}\right]\,\text{d}x=3\ln|x-1|-\frac{8}{x-1}+C\).

\(\frac{3x+5}{(x-1)^2}=\frac{3}{x-1}+\frac{8}{(x-1)^2}\) \(\int\frac{3x+5}{(x-1)^2}\,\text{d}x=3\ln|x-1|-\frac{8}{x-1}+C\)

- Match the decomposition structure to the distinct factors in the denominator. - Once the form is chosen, use each denominator zero separately. - Recombine all three terms to verify the numerator.

1. Each distinct linear factor requires its own constant numerator, so form B is standard. 2. Substituting \(x=0\), \(x=2\), and \(x=-1\) gives \(A=-1\), \(B=1\), and \(C=1\). 3. Thus the integrand is \(-\frac{1}{x}+\frac{1}{x-2}+\frac{1}{x+1}\). 4. The antiderivative is \(-\ln|x|+\ln|x-2|+\ln|x+1|+C\).

Form B, with \(A=-1\), \(B=1\), and \(C=1\). The integral is \(-\ln|x|+\ln|x-2|+\ln|x+1|+C\).

- Include a term for every power of a repeated linear factor. - Use a numerator of degree one less than an irreducible quadratic factor. - This problem asks only for the decomposition structure, not the coefficient values.

1. The repeated linear factor \((x-3)^2\) requires one constant-numerator term for \(x-3\) and one for \((x-3)^2\). 2. The irreducible quadratic factor \(x^2+9\) requires a linear numerator. 3. Therefore, the standard form is \(\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{Cx+D}{x^2+9}\).

\(\frac{A}{x-3}+\frac{B}{(x-3)^2}+\frac{Cx+D}{x^2+9}\)

- Factor the quadratic into the two consecutive linear factors. - The numerator can be reconstructed from two constant partial-fraction coefficients. - Combine the endpoint ratios after applying their coefficients.

1. Factor \(x^2+5x+6=(x+2)(x+3)\). 2. Partial fractions give \(\frac{x+4}{(x+2)(x+3)}=\frac{2}{x+2}-\frac{1}{x+3}\). 3. Endpoint evaluation gives \(2\ln\left(\frac{3}{2}\right)-\ln\left(\frac{4}{3}\right)\). 4. Combining logarithms yields \(\ln\left(\frac{27}{16}\right)\).

\(\ln\left(\frac{27}{16}\right)\)

- Use one constant numerator for each distinct linear factor. - Multiply by the common denominator before solving for the coefficients. - Substituting the zeros of the denominator factors isolates one coefficient at a time.

1. Write \(\frac{3x+7}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}\). 2. Combining the fractions gives \(3x+7=A(x+2)+B(x+1)\). 3. Substitute \(x=-1\) to obtain \(A=4\), and substitute \(x=-2\) to obtain \(B=-1\). 4. Therefore, \(\int\frac{3x+7}{(x+1)(x+2)}\,dx=4\ln|x+1|-\ln|x+2|+C\).

\(4\ln|x+1|-\ln|x+2|+C\)

- Recombine the proposed fractions before deciding whether they are correct. - Values that make the denominator factors zero can isolate the unknown constants. - A sign difference in the decomposition should also appear between the logarithmic terms.

1. Recombining the student's fractions gives \(\frac{2x+1}{3(x-1)(x+2)}\), not the original numerator \(1\). 2. Let \(\frac{1}{(x-1)(x+2)}=\frac{A}{x-1}+\frac{B}{x+2}\). 3. Substitution at the roots gives \(A=\frac{1}{3}\) and \(B=-\frac{1}{3}\). 4. The integral is \(\frac{1}{3}\ln|x-1|-\frac{1}{3}\ln|x+2|+C\).

The second sign should be negative: \(\frac{1}{(x-1)(x+2)}=\frac{1}{3(x-1)}-\frac{1}{3(x+2)}\). The integral is \(\frac{1}{3}\ln|x-1|-\frac{1}{3}\ln|x+2|+C\).

- Include a term for every distinct linear factor, even if one coefficient may later be zero. - Use the three roots to isolate the constants efficiently. - A zero coefficient should be retained as a conclusion, not assumed at the start.

1. Write the decomposition with constants over \(x-1\), \(x+1\), and \(x+2\). 2. Substituting \(x=1\), \(x=-1\), and \(x=-2\) gives \(A=\frac{1}{2}\), \(B=\frac{3}{2}\), and \(C=0\). 3. Thus the integrand is \(\frac{1}{2(x-1)}+\frac{3}{2(x+1)}\). 4. The antiderivative is \(\frac{1}{2}\ln|x-1|+\frac{3}{2}\ln|x+1|+C\).

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

- Decompose the rational function into terms with one linear denominator each. - Use the stated interval to write logarithms with positive arguments. - Apply the point condition only after finding the full antiderivative family.

1. Partial fractions give \(f(x)=\frac{11}{4(x-3)}+\frac{5}{4(x+1)}\). 2. On \((-1,3)\), one antiderivative family is \(F(x)=\frac{11}{4}\ln(3-x)+\frac{5}{4}\ln(x+1)+C\). 3. The condition \(F(0)=2\) gives \(2=\frac{11}{4}\ln3+C\), so \(C=2-\frac{11}{4}\ln3\). 4. Therefore, \(F(x)=2+\frac{11}{4}\ln\left(\frac{3-x}{3}\right)+\frac{5}{4}\ln(x+1)\).

\(F(x)=2+\frac{11}{4}\ln\left(\frac{3-x}{3}\right)+\frac{5}{4}\ln(x+1)\), for \(-1<x<3\)

- Use the zeros of the two linear factors to find the constants. - One denominator has a leading coefficient other than \(1\). - Differentiate each logarithm to confirm its scale factor.

1. Set \(\frac{6x-1}{(2x-1)(x+2)}=\frac{A}{2x-1}+\frac{B}{x+2}\). 2. Substituting \(x=\frac{1}{2}\) and \(x=-2\) gives \(A=\frac{4}{5}\) and \(B=\frac{13}{5}\). 3. Account for the derivative of \(2x-1\) when integrating its term. 4. The result is \(\frac{2}{5}\ln|2x-1|+\frac{13}{5}\ln|x+2|+C\).

\(\frac{2}{5}\ln|2x-1|+\frac{13}{5}\ln|x+2|+C\)

- Write the rational function in terms of the unknown scale factor \(K\). - Use the given function value to determine \(K\) before decomposing. - Substitute the zeros of the two denominator factors to isolate the partial-fraction coefficients.

1. The stated numerator and denominator imply \(R(x)=K\frac{x-1}{(x+3)(x-2)}\). 2. Substitute \(x=0\): \(R(0)=K\frac{-1}{(3)(-2)}=\frac{K}{6}\). Since \(R(0)=\frac{2}{3}\), \(K=4\). 3. Partial fractions give \(R(x)=\frac{16}{5(x+3)}+\frac{4}{5(x-2)}\). 4. Therefore, \(\int R(x)\,\text{d}x=\frac{16}{5}\ln|x+3|+\frac{4}{5}\ln|x-2|+C\).

\(R(x)=\frac{4(x-1)}{(x+3)(x-2)}=\frac{16}{5(x+3)}+\frac{4}{5(x-2)}\) \(\int R(x)\,\text{d}x=\frac{16}{5}\ln|x+3|+\frac{4}{5}\ln|x-2|+C\)

- First express the rational function as a difference of two simpler fractions. - Keep the variable upper endpoint in the logarithmic expression. - Equal logarithms with positive arguments lead to an algebraic equation.

1. Decompose \(\frac{1}{(x+1)(x+2)}=\frac{1}{x+1}-\frac{1}{x+2}\). 2. The integral equals \(\ln\left(\frac{b+1}{b+2}\right)-\ln\left(\frac{1}{2}\right)=\ln\left(\frac{2(b+1)}{b+2}\right)\). 3. Equating logarithm arguments gives \(\frac{2(b+1)}{b+2}=\frac{4}{3}\). 4. Solving yields \(b=1\), which is positive.

\(b=1\)

- Factor the difference of squares before decomposing. - At each endpoint, compare the product of the two absolute-value arguments. - The original integrand also provides a symmetry check.

1. Factor \(x^2-4=(x-2)(x+2)\) and decompose \(\frac{2x}{x^2-4}=\frac{1}{x-2}+\frac{1}{x+2}\). 2. An antiderivative is \(\ln|x-2|+\ln|x+2|\). 3. At both \(x=1\) and \(x=-1\), the product of the absolute-value arguments is \(3\). 4. The endpoint values are equal, so the definite integral is \(0\).

\(\frac{2x}{x^2-4}=\frac{1}{x-2}+\frac{1}{x+2}\), and the definite integral is \(0\). The symmetric endpoints produce equal logarithmic endpoint values.

- The requested net change is the integral of the given rate. - Separate the rate into two terms with one linear denominator each. - Retain the units after evaluating the endpoint logarithms.

1. Decompose \(r(t)=\frac{30}{t+2}-\frac{30}{t+5}\). 2. The net change is \(\int_0^3r(t)\,\text{d}t=30[\ln(t+2)-\ln(t+5)]_0^3\). 3. Endpoint evaluation gives \(30\ln\left(\frac{25}{16}\right)\) megabytes.

\(30\ln\left(\frac{25}{16}\right)\,\text{MB}\)

- Reduce the rational function to a polynomial plus a proper fraction first. - Factor the remainder denominator before introducing constants. - Check the answer by differentiating both the polynomial and logarithmic parts.

1. Polynomial division gives \(\frac{x^3+2x^2-2x+5}{x^2-1}=x+2+\frac{-x+7}{x^2-1}\). 2. Factor the denominator and decompose \(\frac{-x+7}{(x-1)(x+1)}=\frac{3}{x-1}-\frac{4}{x+1}\). 3. Integrating all terms gives \(\frac{x^2}{2}+2x+3\ln|x-1|-4\ln|x+1|+C\).

\(\frac{x^2}{2}+2x+3\ln|x-1|-4\ln|x+1|+C\)

- Decompose the rational function before considering its domain. - The denominator zeros divide the real line into separate intervals. - An arbitrary constant connects antiderivatives only within one connected domain interval.

1. Partial fractions give \(\frac{2x+3}{(x-1)(x+2)}=\frac{5}{3(x-1)}+\frac{1}{3(x+2)}\). 2. An antiderivative is \(\frac{5}{3}\ln|x-1|+\frac{1}{3}\ln|x+2|+C\). 3. The domain intervals are \((-\infty, -2)\), \((-2, 1)\), and \((1, \infty)\). 4. The rational function is undefined at \(-2\) and \(1\), so an antiderivative on one interval is not connected to an antiderivative on another; each interval may have its own constant.

\(\frac{5}{3}\ln|x-1|+\frac{1}{3}\ln|x+2|+C\), separately on \((-\infty, -2)\), \((-2, 1)\), and \((1, \infty)\). The constant may differ from interval to interval.

- Recombine the student's fractions to inspect the resulting numerator. - Each root isolates the coefficient attached to the other factor. - The corrected coefficients should reproduce both the variable and constant terms.

1. The proposed right side recombines to \(\frac{(x+1)+3(x-2)}{(x-2)(x+1)}=\frac{4x-5}{(x-2)(x+1)}\). 2. Substituting \(x=2\) in the correct setup gives the coefficient over \(x-2\) as \(3\). 3. Substituting \(x=-1\) gives the coefficient over \(x+1\) as \(1\). 4. Therefore the integral is \(3\ln|x-2|+\ln|x+1|+C\).

The correct decomposition is \(\frac{4x+1}{(x-2)(x+1)}=\frac{3}{x-2}+\frac{1}{x+1}\). The integral is \(3\ln|x-2|+\ln|x+1|+C\).

- Use each of the three roots to obtain one coefficient. - The interval keeps all factor values positive. - Combine half-coefficient logarithms carefully at the end.

1. Decompose \(\frac{1}{(x+1)(x+2)(x+3)}=\frac{1}{2(x+1)}-\frac{1}{x+2}+\frac{1}{2(x+3)}\). 2. An antiderivative is \(\frac{1}{2}\ln(x+1)-\ln(x+2)+\frac{1}{2}\ln(x+3)\) on \([0, 1]\). 3. Evaluating the endpoints and combining logarithms gives \(\frac{1}{2}\ln\left(\frac{32}{27}\right)\).

\(\frac{1}{2}\ln\left(\frac{32}{27}\right)\)

- Clear the denominator and expand the numerator identity. - Compare the variable coefficient and constant term separately. - Solve the resulting two-equation system before integrating.

1. Set \(4x+7=A(x+3)+B(x-1)\). 2. Matching coefficients gives \(A+B=4\) and \(3A-B=7\). 3. Solving gives \(A=\frac{11}{4}\) and \(B=\frac{5}{4}\). 4. The antiderivative is \(\frac{11}{4}\ln|x-1|+\frac{5}{4}\ln|x+3|+C\).

\(\frac{4x+7}{(x-1)(x+3)}=\frac{11}{4(x-1)}+\frac{5}{4(x+3)}\) \(\frac{11}{4}\ln|x-1|+\frac{5}{4}\ln|x+3|+C\)

- Clear the denominator and compare the coefficient of \(x\) on both sides. - One denominator root isolates one coefficient directly. - Use the coefficient-sum equation to find the remaining value.

1. Clear the denominators: \(5x+2=A(x+4)+B(x-1)\). 2. Comparing the coefficients of \(x\) gives \(A+B=5\). 3. Substituting \(x=1\) gives \(7=5A\), so \(A=\frac{7}{5}\). 4. Therefore, \(B=5-\frac{7}{5}=\frac{18}{5}\). 5. The antiderivative is \(\frac{7}{5}\ln|x-1|+\frac{18}{5}\ln|x+4|+C\).

\(A=\frac{7}{5}\), \(B=\frac{18}{5}\) \(\frac{7}{5}\ln|x-1|+\frac{18}{5}\ln|x+4|+C\)

- Multiply by the factor associated with one denominator zero before taking the matching limit. - Repeat the same coefficient calculation at each denominator zero. - Recombine the three terms if any sign is uncertain.

1. The coefficient at \(x=2\) is \(A=\frac{2^2+1}{(2+1)(2+3)}=\frac{1}{3}\). 2. The coefficients at \(x=-1\) and \(x=-3\) are \(B=-\frac{1}{3}\) and \(C=1\), respectively. 3. Thus \(R(x)=\frac{1}{3(x-2)}-\frac{1}{3(x+1)}+\frac{1}{x+3}\). 4. The antiderivative is \(\frac{1}{3}\ln|x-2|-\frac{1}{3}\ln|x+1|+\ln|x+3|+C\).

\(R(x)=\frac{1}{3(x-2)}-\frac{1}{3(x+1)}+\frac{1}{x+3}\) \(\int R(x)\,\text{d}x=\frac{1}{3}\ln|x-2|-\frac{1}{3}\ln|x+1|+\ln|x+3|+C\)

- Use a constant numerator over the linear factor and a linear numerator over the irreducible quadratic. - Match coefficients after recombining the fractions. - Split the quadratic term into a denominator-derivative part and an inverse-tangent part.

1. Use the form \(\frac{2x+7}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4}\). 2. Recombining gives \(2x+7=A(x^2+4)+(Bx+C)(x+1)\). 3. Matching coefficients yields \(A=1\), \(B=-1\), and \(C=3\). 4. Integrate term by term: \(\ln|x+1|-\frac{1}{2}\ln(x^2+4)+\frac{3}{2}\arctan\left(\frac{x}{2}\right)+C\).

\(\frac{2x+7}{(x+1)(x^2+4)}=\frac{1}{x+1}+\frac{-x+3}{x^2+4}\) \(\ln|x+1|-\frac{1}{2}\ln(x^2+4)+\frac{3}{2}\arctan\left(\frac{x}{2}\right)+C\)

- Total output is the integral of the processing rate over time. - Separate the rational rate into a difference of two simple fractions. - Retain the task unit after evaluating the endpoint logarithms.

1. Decompose \(r(t)=\frac{3}{t+1}-\frac{3}{t+3}\). 2. The total number of tasks is \(\int_0^2r(t)\,\text{d}t=3[\ln(t+1)-\ln(t+3)]_0^2\). 3. Endpoint evaluation gives \(3\ln\left(\frac{9}{5}\right)\) tasks.

\(3\ln\left(\frac{9}{5}\right)\) tasks

- Use each denominator zero to isolate one coefficient. - Integrate the three simple fractions before combining anything. - Positive logarithmic coefficients become numerator powers, while negative ones become denominator powers.

1. Substituting the three denominator zeros gives coefficients \(2\), \(-1\), and \(-1\). 2. The integrand is \(\frac{2}{x-1}-\frac{1}{x+2}-\frac{1}{x-4}\). 3. The antiderivative is \(2\ln|x-1|-\ln|x+2|-\ln|x-4|+C\). 4. Combining logarithms gives \(\ln\left|\frac{(x-1)^2}{(x+2)(x-4)}\right|+C\).

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

- Expand the requested single logarithm into a sum of two logarithms. - Equal logarithmic coefficients mean equal partial-fraction coefficients. - Recombine those equal fractions and match the variable and constant terms.

1. The requested logarithm expands to \(c\ln|x-1|+c\ln|x+2|\), so the two partial-fraction coefficients must both equal \(c\). 2. Recombine: \(\frac{c}{x-1}+\frac{c}{x+2}=\frac{c(2x+1)}{(x-1)(x+2)}\). 3. Match \(c(2x+1)\) with \(4x+b\). The x-coefficient gives \(2c=4\), so \(c=2\). 4. The constant term then gives \(b=c=2\).

\(b=2\) and \(c=2\)

- Use the difference-of-squares factorization without changing the first factor’s orientation. - Find one constant over each linear factor. - One factor has derivative \(-1\), which affects its logarithmic sign.

1. Factor \(4-x^2=(2-x)(2+x)\). 2. Decompose \(\frac{x+1}{(2-x)(2+x)}=\frac{3}{4(2-x)}-\frac{1}{4(2+x)}\). 3. Since the derivative of \(2-x\) is \(-1\), its logarithmic term gains another negative sign. 4. The antiderivative is \(-\frac{3}{4}\ln|2-x|-\frac{1}{4}\ln|2+x|+C\).

\(-\frac{3}{4}\ln|2-x|-\frac{1}{4}\ln|2+x|+C\)

- Translate each table row directly into one simple fraction. - Use the product of all three factors as the common denominator. - Keep the decomposed form for the integration step.

1. The decomposed function is \(\frac{1}{x+2}-\frac{2}{x-1}+\frac{3}{x-4}\). 2. Over the common denominator, the numerator is \((x-1)(x-4)-2(x+2)(x-4)+3(x+2)(x-1)=2x^{2}+2x+14\). 3. The rational function is \(\frac{2x^{2}+2x+14}{(x+2)(x-1)(x-4)}\). 4. Its antiderivative is \(\ln|x+2|-2\ln|x-1|+3\ln|x-4|+C\).

\(\frac{2x^{2}+2x+14}{(x+2)(x-1)(x-4)}\) \(\ln|x+2|-2\ln|x-1|+3\ln|x-4|+C\)

- The total amount is the integral of the rate over the stated minute. - Decompose the rational factor before carrying the parameter through the integral. - Solve for the positive scale parameter after evaluating the logarithmic change.

1. Decompose \(\frac{1}{(t+1)(t+2)}=\frac{1}{t+1}-\frac{1}{t+2}\). 2. The accumulated volume is \(k[\ln(t+1)-\ln(t+2)]_0^1=k\ln\left(\frac{4}{3}\right)\). 3. Set this equal to \(10\): \(k\ln\left(\frac{4}{3}\right)=10\). 4. Therefore \(k=\frac{10}{\ln(4/3)}\,\text{mL/min}\).

\(k=\frac{10}{\ln(4/3)}\,\text{mL/min}\)

- Use the zero of \(x-4\) to express its coefficient directly in terms of \(p\). - After determining \(p\), use the other two denominator zeros to obtain their coefficients. - Integrate each simple fraction logarithmically.

1. The coefficient over \(x-4\) is found by substituting \(x=4\) after canceling that factor: \(C=\frac{16+4p-6}{(4-1)(4+2)}=\frac{10+4p}{18}\). 2. Set \(C=-1\): \(10+4p=-18\), so \(p=-7\). 3. With \(p=-7\), substitute the other denominator zeros. At \(x=1\), \(A=\frac{1-7-6}{(1+2)(1-4)}=\frac{4}{3}\). At \(x=-2\), \(B=\frac{4+14-6}{(-2-1)(-2-4)}=\frac{2}{3}\). 4. Therefore, \(R(x)=\frac{4}{3(x-1)}+\frac{2}{3(x+2)}-\frac{1}{x-4}\). 5. The antiderivative is \(\frac{4}{3}\ln|x-1|+\frac{2}{3}\ln|x+2|-\ln|x-4|+C\).

\(p=-7\) \(R(x)=\frac{4}{3(x-1)}+\frac{2}{3(x+2)}-\frac{1}{x-4}\) \(\int R(x)\,\text{d}x=\frac{4}{3}\ln|x-1|+\frac{2}{3}\ln|x+2|-\ln|x-4|+C\)

- Write both partial-fraction coefficients as expressions involving the parameter. - Evaluate the two simple definite integrals before solving. - The resulting equation is linear in the parameter even though it contains logarithms.

1. Decompose \(\frac{kx+3}{(x+1)(x+2)}=\frac{3-k}{x+1}+\frac{2k-3}{x+2}\). 2. Integrating from \(0\) to \(1\) gives \((3-k)\ln(2)+(2k-3)\ln\left(\frac{3}{2}\right)\). 3. Set this equal to \(\ln(3)\). Substituting \(k=2\) gives \(\ln(2)+\ln\left(\frac{3}{2}\right)=\ln(3)\). 4. The left side is linear in \(k\) with nonzero coefficient \(\ln\left(\frac{9}{8}\right)\), so this solution is unique.

\(k=2\)

- Compare the end behavior of the original expression and the proposed decomposition. - Perform division before decomposing the remainder. - Only the proper fraction should be assigned simple linear denominators.

1. The student's right side approaches \(0\) as \(|x|\) grows, but the original rational function approaches \(2\), so a polynomial term is missing. 2. Division gives \(\frac{2x^2+x+4}{x^2+x-2}=2+\frac{-x+8}{(x-1)(x+2)}\). 3. The proper remainder decomposes as \(\frac{-x+8}{(x-1)(x+2)}=\frac{7}{3(x-1)}-\frac{10}{3(x+2)}\). 4. The integral is \(2x+\frac{7}{3}\ln|x-1|-\frac{10}{3}\ln|x+2|+C\).

A constant polynomial term is required: \(\frac{2x^2+x+4}{(x-1)(x+2)}=2+\frac{7}{3(x-1)}-\frac{10}{3(x+2)}\). The integral is \(2x+\frac{7}{3}\ln|x-1|-\frac{10}{3}\ln|x+2|+C\).

- Clear the denominator and compare the leading coefficient first. - Use the given product together with the coefficient sum to find the unordered pair. - The constant term distinguishes the two possible assignments.

1. Matching the coefficient of \(x\) gives \(A+B=5\). 2. Together with \(AB=6\), the partial-fraction coefficients are the roots of \(z^2-5z+6=0\), so \(\{A, B\}=\{2, 3\}\). 3. The constant term satisfies \(k=3A-2B\). 4. If \((A, B)=(2, 3)\), then \(k=0\), and the integral is \(2\ln|x-2|+3\ln|x+3|+C\). 5. If \((A, B)=(3, 2)\), then \(k=5\), and the integral is \(3\ln|x-2|+2\ln|x+3|+C\).

\(k=0\): \(A=2\), \(B=3\), and the integral is \(2\ln|x-2|+3\ln|x+3|+C\). \(k=5\): \(A=3\), \(B=2\), and the integral is \(3\ln|x-2|+2\ln|x+3|+C\).