Hard A-Level Integration with partial fractions Questions

Challenging, exam-style A-Level Integration with partial fractions questions with worked solutions. Stretch yourself on the hardest partial fractions, integration problems.

partial fractionsintegration
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Which of the following is equal to ∫3x2+1(x−4)(x−1)(x+2) dx\int \frac{3 x^{2} + 1}{\left(x - 4\right) \left(x - 1\right) \left(x + 2\right)}\,dx?
Show worked solution

Worked solution

  1. Write down the integral

    ∫3x2+1(x−4)(x−1)(x+2) dx\int \frac{3 x^{2} + 1}{\left(x - 4\right) \left(x - 1\right) \left(x + 2\right)}\,dx

    We integrate this rational function by first splitting it into partial fractions.

  2. Factorise the denominator

    x3−3x2−6x+8=(x−4)(x−1)(x+2)x^{3} - 3 x^{2} - 6 x + 8=\left(x - 4\right) \left(x - 1\right) \left(x + 2\right)

    The linear factors tell us the form of the partial fractions.

  3. Set up the partial fraction form

    3x2+1(x−4)(x−1)(x+2)≡Ax+2+Bx−1+Cx−4\frac{3 x^{2} + 1}{\left(x - 4\right) \left(x - 1\right) \left(x + 2\right)}\equiv \frac{A}{x + 2}+\frac{B}{x - 1}+\frac{C}{x - 4}

    Write one constant over each distinct linear factor.

  4. Multiply through by the denominator

    3x2+1≡A((x−4)(x−1))+B((x−4)(x+2))+C((x−1)(x+2))3 x^{2} + 1\equiv A\left(\left(x - 4\right) \left(x - 1\right)\right)+B\left(\left(x - 4\right) \left(x + 2\right)\right)+C\left(\left(x - 1\right) \left(x + 2\right)\right)

    Clearing the fractions gives an identity valid for all x.

  5. Substitute x=-2 to find A

    A=1318A=\frac{13}{18}

    This choice of x makes the other unknowns vanish (cover-up method).

  6. Substitute x=1 to find B

    B=−49B=- \frac{4}{9}

    This choice of x makes the other unknowns vanish (cover-up method).

  7. Substitute x=4 to find C

    C=4918C=\frac{49}{18}

    This choice of x makes the other unknowns vanish (cover-up method).

  8. Collect the constants

    A=1318, B=−49, C=4918A=\frac{13}{18},\ B=- \frac{4}{9},\ C=\frac{49}{18}

    These constants complete the decomposition.

  9. Write the partial fraction decomposition

    3x2+1(x−4)(x−1)(x+2)≡1318(x+2)−49(x−1)+4918(x−4)\frac{3 x^{2} + 1}{\left(x - 4\right) \left(x - 1\right) \left(x + 2\right)}\equiv \frac{13}{18 \left(x + 2\right)} - \frac{4}{9 \left(x - 1\right)} + \frac{49}{18 \left(x - 4\right)}

    Each simple fraction is now easy to integrate.

  10. Split into standard integrals

    ∫3x2+1(x−4)(x−1)(x+2) dx=∫1318(x+2) dx+∫−49(x−1) dx+∫4918(x−4) dx\int \frac{3 x^{2} + 1}{\left(x - 4\right) \left(x - 1\right) \left(x + 2\right)}\,dx=\int \frac{13}{18 \left(x + 2\right)}\,dx+\int - \frac{4}{9 \left(x - 1\right)}\,dx+\int \frac{49}{18 \left(x - 4\right)}\,dx

    Each term of the form k/(x-a) integrates to a logarithm.

  11. Integrate the first term

    ∫1318(x+2) dx=13ln⁡(18x+36)18\int \frac{13}{18 \left(x + 2\right)}\,dx=\frac{13 \ln{\left(18 x + 36 \right)}}{18}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  12. Integrate the second term

    ∫−49(x−1) dx=−4ln⁡(9x−9)9\int - \frac{4}{9 \left(x - 1\right)}\,dx=- \frac{4 \ln{\left(9 x - 9 \right)}}{9}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  13. Integrate the third term

    ∫4918(x−4) dx=49ln⁡(18x−72)18\int \frac{49}{18 \left(x - 4\right)}\,dx=\frac{49 \ln{\left(18 x - 72 \right)}}{18}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  14. Combine the integrated terms

    49ln⁡(x−4)18−4ln⁡(x−1)9+13ln⁡(x+2)18\frac{49 \ln{\left(x - 4 \right)}}{18} - \frac{4 \ln{\left(x - 1 \right)}}{9} + \frac{13 \ln{\left(x + 2 \right)}}{18}

    Add the logarithmic terms to form the antiderivative.

  15. Select the correct option

    49ln⁡(x−4)18−4ln⁡(x−1)9+13ln⁡(x+2)18+c\frac{49 \ln{\left(x - 4 \right)}}{18} - \frac{4 \ln{\left(x - 1 \right)}}{9} + \frac{13 \ln{\left(x + 2 \right)}}{18} + c

    The option matching the integrated decomposition is the correct answer.

Answer
49ln⁡(x−4)18−4ln⁡(x−1)9+13ln⁡(x+2)18+c\frac{49 \ln{\left(x - 4 \right)}}{18} - \frac{4 \ln{\left(x - 1 \right)}}{9} + \frac{13 \ln{\left(x + 2 \right)}}{18} + c
Question 2
8 markschallenging
Which of the following is equal to ∫x2+5(x−3)(x−1)(x+2) dx\int \frac{x^{2} + 5}{\left(x - 3\right) \left(x - 1\right) \left(x + 2\right)}\,dx?
Show worked solution

Worked solution

  1. Write down the integral

    ∫x2+5(x−3)(x−1)(x+2) dx\int \frac{x^{2} + 5}{\left(x - 3\right) \left(x - 1\right) \left(x + 2\right)}\,dx

    We integrate this rational function by first splitting it into partial fractions.

  2. Factorise the denominator

    x3−2x2−5x+6=(x−3)(x−1)(x+2)x^{3} - 2 x^{2} - 5 x + 6=\left(x - 3\right) \left(x - 1\right) \left(x + 2\right)

    The linear factors tell us the form of the partial fractions.

  3. Set up the partial fraction form

    x2+5(x−3)(x−1)(x+2)≡Ax+2+Bx−1+Cx−3\frac{x^{2} + 5}{\left(x - 3\right) \left(x - 1\right) \left(x + 2\right)}\equiv \frac{A}{x + 2}+\frac{B}{x - 1}+\frac{C}{x - 3}

    Write one constant over each distinct linear factor.

  4. Multiply through by the denominator

    x2+5≡A((x−3)(x−1))+B((x−3)(x+2))+C((x−1)(x+2))x^{2} + 5\equiv A\left(\left(x - 3\right) \left(x - 1\right)\right)+B\left(\left(x - 3\right) \left(x + 2\right)\right)+C\left(\left(x - 1\right) \left(x + 2\right)\right)

    Clearing the fractions gives an identity valid for all x.

  5. Substitute x=-2 to find A

    A=35A=\frac{3}{5}

    This choice of x makes the other unknowns vanish (cover-up method).

  6. Substitute x=1 to find B

    B=−1B=-1

    This choice of x makes the other unknowns vanish (cover-up method).

  7. Substitute x=3 to find C

    C=75C=\frac{7}{5}

    This choice of x makes the other unknowns vanish (cover-up method).

  8. Collect the constants

    A=35, B=−1, C=75A=\frac{3}{5},\ B=-1,\ C=\frac{7}{5}

    These constants complete the decomposition.

  9. Write the partial fraction decomposition

    x2+5(x−3)(x−1)(x+2)≡35(x+2)−1x−1+75(x−3)\frac{x^{2} + 5}{\left(x - 3\right) \left(x - 1\right) \left(x + 2\right)}\equiv \frac{3}{5 \left(x + 2\right)} - \frac{1}{x - 1} + \frac{7}{5 \left(x - 3\right)}

    Each simple fraction is now easy to integrate.

  10. Split into standard integrals

    ∫x2+5(x−3)(x−1)(x+2) dx=∫35(x+2) dx+∫−1x−1 dx+∫75(x−3) dx\int \frac{x^{2} + 5}{\left(x - 3\right) \left(x - 1\right) \left(x + 2\right)}\,dx=\int \frac{3}{5 \left(x + 2\right)}\,dx+\int - \frac{1}{x - 1}\,dx+\int \frac{7}{5 \left(x - 3\right)}\,dx

    Each term of the form k/(x-a) integrates to a logarithm.

  11. Integrate the first term

    ∫35(x+2) dx=3ln⁡(5x+10)5\int \frac{3}{5 \left(x + 2\right)}\,dx=\frac{3 \ln{\left(5 x + 10 \right)}}{5}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  12. Integrate the second term

    ∫−1x−1 dx=−ln⁡(x−1)\int - \frac{1}{x - 1}\,dx=- \ln{\left(x - 1 \right)}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  13. Integrate the third term

    ∫75(x−3) dx=7ln⁡(5x−15)5\int \frac{7}{5 \left(x - 3\right)}\,dx=\frac{7 \ln{\left(5 x - 15 \right)}}{5}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  14. Combine the integrated terms

    7ln⁡(x−3)5−ln⁡(x−1)+3ln⁡(x+2)5\frac{7 \ln{\left(x - 3 \right)}}{5} - \ln{\left(x - 1 \right)} + \frac{3 \ln{\left(x + 2 \right)}}{5}

    Add the logarithmic terms to form the antiderivative.

  15. Select the correct option

    7ln⁡(x−3)5−ln⁡(x−1)+3ln⁡(x+2)5+c\frac{7 \ln{\left(x - 3 \right)}}{5} - \ln{\left(x - 1 \right)} + \frac{3 \ln{\left(x + 2 \right)}}{5} + c

    The option matching the integrated decomposition is the correct answer.

Answer
7ln⁡(x−3)5−ln⁡(x−1)+3ln⁡(x+2)5+c\frac{7 \ln{\left(x - 3 \right)}}{5} - \ln{\left(x - 1 \right)} + \frac{3 \ln{\left(x + 2 \right)}}{5} + c
Question 3
8 markschallenging
Which of the following is equal to ∫x2+x+1(x−2)(x−1)(x+1) dx\int \frac{x^{2} + x + 1}{\left(x - 2\right) \left(x - 1\right) \left(x + 1\right)}\,dx?
Show worked solution

Worked solution

  1. Write down the integral

    ∫x2+x+1(x−2)(x−1)(x+1) dx\int \frac{x^{2} + x + 1}{\left(x - 2\right) \left(x - 1\right) \left(x + 1\right)}\,dx

    We integrate this rational function by first splitting it into partial fractions.

  2. Factorise the denominator

    x3−2x2−x+2=(x−2)(x−1)(x+1)x^{3} - 2 x^{2} - x + 2=\left(x - 2\right) \left(x - 1\right) \left(x + 1\right)

    The linear factors tell us the form of the partial fractions.

  3. Set up the partial fraction form

    x2+x+1(x−2)(x−1)(x+1)≡Ax+1+Bx−1+Cx−2\frac{x^{2} + x + 1}{\left(x - 2\right) \left(x - 1\right) \left(x + 1\right)}\equiv \frac{A}{x + 1}+\frac{B}{x - 1}+\frac{C}{x - 2}

    Write one constant over each distinct linear factor.

  4. Multiply through by the denominator

    x2+x+1≡A((x−2)(x−1))+B((x−2)(x+1))+C((x−1)(x+1))x^{2} + x + 1\equiv A\left(\left(x - 2\right) \left(x - 1\right)\right)+B\left(\left(x - 2\right) \left(x + 1\right)\right)+C\left(\left(x - 1\right) \left(x + 1\right)\right)

    Clearing the fractions gives an identity valid for all x.

  5. Substitute x=-1 to find A

    A=16A=\frac{1}{6}

    This choice of x makes the other unknowns vanish (cover-up method).

  6. Substitute x=1 to find B

    B=−32B=- \frac{3}{2}

    This choice of x makes the other unknowns vanish (cover-up method).

  7. Substitute x=2 to find C

    C=73C=\frac{7}{3}

    This choice of x makes the other unknowns vanish (cover-up method).

  8. Collect the constants

    A=16, B=−32, C=73A=\frac{1}{6},\ B=- \frac{3}{2},\ C=\frac{7}{3}

    These constants complete the decomposition.

  9. Write the partial fraction decomposition

    x2+x+1(x−2)(x−1)(x+1)≡16(x+1)−32(x−1)+73(x−2)\frac{x^{2} + x + 1}{\left(x - 2\right) \left(x - 1\right) \left(x + 1\right)}\equiv \frac{1}{6 \left(x + 1\right)} - \frac{3}{2 \left(x - 1\right)} + \frac{7}{3 \left(x - 2\right)}

    Each simple fraction is now easy to integrate.

  10. Split into standard integrals

    ∫x2+x+1(x−2)(x−1)(x+1) dx=∫16(x+1) dx+∫−32(x−1) dx+∫73(x−2) dx\int \frac{x^{2} + x + 1}{\left(x - 2\right) \left(x - 1\right) \left(x + 1\right)}\,dx=\int \frac{1}{6 \left(x + 1\right)}\,dx+\int - \frac{3}{2 \left(x - 1\right)}\,dx+\int \frac{7}{3 \left(x - 2\right)}\,dx

    Each term of the form k/(x-a) integrates to a logarithm.

  11. Integrate the first term

    ∫16(x+1) dx=ln⁡(6x+6)6\int \frac{1}{6 \left(x + 1\right)}\,dx=\frac{\ln{\left(6 x + 6 \right)}}{6}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  12. Integrate the second term

    ∫−32(x−1) dx=−3ln⁡(2x−2)2\int - \frac{3}{2 \left(x - 1\right)}\,dx=- \frac{3 \ln{\left(2 x - 2 \right)}}{2}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  13. Integrate the third term

    ∫73(x−2) dx=7ln⁡(3x−6)3\int \frac{7}{3 \left(x - 2\right)}\,dx=\frac{7 \ln{\left(3 x - 6 \right)}}{3}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  14. Combine the integrated terms

    7ln⁡(x−2)3−3ln⁡(x−1)2+ln⁡(x+1)6\frac{7 \ln{\left(x - 2 \right)}}{3} - \frac{3 \ln{\left(x - 1 \right)}}{2} + \frac{\ln{\left(x + 1 \right)}}{6}

    Add the logarithmic terms to form the antiderivative.

  15. Select the correct option

    7ln⁡(x−2)3−3ln⁡(x−1)2+ln⁡(x+1)6+c\frac{7 \ln{\left(x - 2 \right)}}{3} - \frac{3 \ln{\left(x - 1 \right)}}{2} + \frac{\ln{\left(x + 1 \right)}}{6} + c

    The option matching the integrated decomposition is the correct answer.

Answer
7ln⁡(x−2)3−3ln⁡(x−1)2+ln⁡(x+1)6+c\frac{7 \ln{\left(x - 2 \right)}}{3} - \frac{3 \ln{\left(x - 1 \right)}}{2} + \frac{\ln{\left(x + 1 \right)}}{6} + c
Question 4
8 markschallenging
The expression 2x2+1(x−3)(x−2)(x+1)\frac{2 x^{2} + 1}{\left(x - 3\right) \left(x - 2\right) \left(x + 1\right)} is written in partial fractions. Which option is correct?
Show worked solution

Worked solution

  1. Write down the integral

    ∫2x2+1(x−3)(x−2)(x+1) dx\int \frac{2 x^{2} + 1}{\left(x - 3\right) \left(x - 2\right) \left(x + 1\right)}\,dx

    We integrate this rational function by first splitting it into partial fractions.

  2. Factorise the denominator

    x3−4x2+x+6=(x−3)(x−2)(x+1)x^{3} - 4 x^{2} + x + 6=\left(x - 3\right) \left(x - 2\right) \left(x + 1\right)

    The linear factors tell us the form of the partial fractions.

  3. Set up the partial fraction form

    2x2+1(x−3)(x−2)(x+1)≡Ax+1+Bx−2+Cx−3\frac{2 x^{2} + 1}{\left(x - 3\right) \left(x - 2\right) \left(x + 1\right)}\equiv \frac{A}{x + 1}+\frac{B}{x - 2}+\frac{C}{x - 3}

    Write one constant over each distinct linear factor.

  4. Multiply through by the denominator

    2x2+1≡A((x−3)(x−2))+B((x−3)(x+1))+C((x−2)(x+1))2 x^{2} + 1\equiv A\left(\left(x - 3\right) \left(x - 2\right)\right)+B\left(\left(x - 3\right) \left(x + 1\right)\right)+C\left(\left(x - 2\right) \left(x + 1\right)\right)

    Clearing the fractions gives an identity valid for all x.

  5. Substitute x=-1 to find A

    A=14A=\frac{1}{4}

    This choice of x makes the other unknowns vanish (cover-up method).

  6. Substitute x=2 to find B

    B=−3B=-3

    This choice of x makes the other unknowns vanish (cover-up method).

  7. Substitute x=3 to find C

    C=194C=\frac{19}{4}

    This choice of x makes the other unknowns vanish (cover-up method).

  8. Collect the constants

    A=14, B=−3, C=194A=\frac{1}{4},\ B=-3,\ C=\frac{19}{4}

    These constants complete the decomposition.

  9. Write the partial fraction decomposition

    2x2+1(x−3)(x−2)(x+1)≡14(x+1)−3x−2+194(x−3)\frac{2 x^{2} + 1}{\left(x - 3\right) \left(x - 2\right) \left(x + 1\right)}\equiv \frac{1}{4 \left(x + 1\right)} - \frac{3}{x - 2} + \frac{19}{4 \left(x - 3\right)}

    Each simple fraction is now easy to integrate.

  10. Split into standard integrals

    ∫2x2+1(x−3)(x−2)(x+1) dx=∫14(x+1) dx+∫−3x−2 dx+∫194(x−3) dx\int \frac{2 x^{2} + 1}{\left(x - 3\right) \left(x - 2\right) \left(x + 1\right)}\,dx=\int \frac{1}{4 \left(x + 1\right)}\,dx+\int - \frac{3}{x - 2}\,dx+\int \frac{19}{4 \left(x - 3\right)}\,dx

    Each term of the form k/(x-a) integrates to a logarithm.

  11. Integrate the first term

    ∫14(x+1) dx=ln⁡(4x+4)4\int \frac{1}{4 \left(x + 1\right)}\,dx=\frac{\ln{\left(4 x + 4 \right)}}{4}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  12. Integrate the second term

    ∫−3x−2 dx=−3ln⁡(x−2)\int - \frac{3}{x - 2}\,dx=- 3 \ln{\left(x - 2 \right)}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  13. Integrate the third term

    ∫194(x−3) dx=19ln⁡(4x−12)4\int \frac{19}{4 \left(x - 3\right)}\,dx=\frac{19 \ln{\left(4 x - 12 \right)}}{4}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  14. Combine the integrated terms

    19ln⁡(x−3)4−3ln⁡(x−2)+ln⁡(x+1)4\frac{19 \ln{\left(x - 3 \right)}}{4} - 3 \ln{\left(x - 2 \right)} + \frac{\ln{\left(x + 1 \right)}}{4}

    Add the logarithmic terms to form the antiderivative.

  15. Select the correct option

    14(x+1)−3x−2+194(x−3)\frac{1}{4 \left(x + 1\right)} - \frac{3}{x - 2} + \frac{19}{4 \left(x - 3\right)}

    The option matching the constants found is the correct answer.

Answer
14(x+1)−3x−2+194(x−3)\frac{1}{4 \left(x + 1\right)} - \frac{3}{x - 2} + \frac{19}{4 \left(x - 3\right)}
Question 5
8 markschallenging
The expression x2+2(x−3)(x−1)(x+1)\frac{x^{2} + 2}{\left(x - 3\right) \left(x - 1\right) \left(x + 1\right)} is written in partial fractions. Which option is correct?
Show worked solution

Worked solution

  1. Write down the integral

    ∫x2+2(x−3)(x−1)(x+1) dx\int \frac{x^{2} + 2}{\left(x - 3\right) \left(x - 1\right) \left(x + 1\right)}\,dx

    We integrate this rational function by first splitting it into partial fractions.

  2. Factorise the denominator

    x3−3x2−x+3=(x−3)(x−1)(x+1)x^{3} - 3 x^{2} - x + 3=\left(x - 3\right) \left(x - 1\right) \left(x + 1\right)

    The linear factors tell us the form of the partial fractions.

  3. Set up the partial fraction form

    x2+2(x−3)(x−1)(x+1)≡Ax+1+Bx−1+Cx−3\frac{x^{2} + 2}{\left(x - 3\right) \left(x - 1\right) \left(x + 1\right)}\equiv \frac{A}{x + 1}+\frac{B}{x - 1}+\frac{C}{x - 3}

    Write one constant over each distinct linear factor.

  4. Multiply through by the denominator

    x2+2≡A((x−3)(x−1))+B((x−3)(x+1))+C((x−1)(x+1))x^{2} + 2\equiv A\left(\left(x - 3\right) \left(x - 1\right)\right)+B\left(\left(x - 3\right) \left(x + 1\right)\right)+C\left(\left(x - 1\right) \left(x + 1\right)\right)

    Clearing the fractions gives an identity valid for all x.

  5. Substitute x=-1 to find A

    A=38A=\frac{3}{8}

    This choice of x makes the other unknowns vanish (cover-up method).

  6. Substitute x=1 to find B

    B=−34B=- \frac{3}{4}

    This choice of x makes the other unknowns vanish (cover-up method).

  7. Substitute x=3 to find C

    C=118C=\frac{11}{8}

    This choice of x makes the other unknowns vanish (cover-up method).

  8. Collect the constants

    A=38, B=−34, C=118A=\frac{3}{8},\ B=- \frac{3}{4},\ C=\frac{11}{8}

    These constants complete the decomposition.

  9. Write the partial fraction decomposition

    x2+2(x−3)(x−1)(x+1)≡38(x+1)−34(x−1)+118(x−3)\frac{x^{2} + 2}{\left(x - 3\right) \left(x - 1\right) \left(x + 1\right)}\equiv \frac{3}{8 \left(x + 1\right)} - \frac{3}{4 \left(x - 1\right)} + \frac{11}{8 \left(x - 3\right)}

    Each simple fraction is now easy to integrate.

  10. Split into standard integrals

    ∫x2+2(x−3)(x−1)(x+1) dx=∫38(x+1) dx+∫−34(x−1) dx+∫118(x−3) dx\int \frac{x^{2} + 2}{\left(x - 3\right) \left(x - 1\right) \left(x + 1\right)}\,dx=\int \frac{3}{8 \left(x + 1\right)}\,dx+\int - \frac{3}{4 \left(x - 1\right)}\,dx+\int \frac{11}{8 \left(x - 3\right)}\,dx

    Each term of the form k/(x-a) integrates to a logarithm.

  11. Integrate the first term

    ∫38(x+1) dx=3ln⁡(8x+8)8\int \frac{3}{8 \left(x + 1\right)}\,dx=\frac{3 \ln{\left(8 x + 8 \right)}}{8}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  12. Integrate the second term

    ∫−34(x−1) dx=−3ln⁡(4x−4)4\int - \frac{3}{4 \left(x - 1\right)}\,dx=- \frac{3 \ln{\left(4 x - 4 \right)}}{4}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  13. Integrate the third term

    ∫118(x−3) dx=11ln⁡(8x−24)8\int \frac{11}{8 \left(x - 3\right)}\,dx=\frac{11 \ln{\left(8 x - 24 \right)}}{8}

    Use \int \frac{k}{x-a}\,dx=k\ln|x-a|.

  14. Combine the integrated terms

    11ln⁡(x−3)8−3ln⁡(x−1)4+3ln⁡(x+1)8\frac{11 \ln{\left(x - 3 \right)}}{8} - \frac{3 \ln{\left(x - 1 \right)}}{4} + \frac{3 \ln{\left(x + 1 \right)}}{8}

    Add the logarithmic terms to form the antiderivative.

  15. Select the correct option

    38(x+1)−34(x−1)+118(x−3)\frac{3}{8 \left(x + 1\right)} - \frac{3}{4 \left(x - 1\right)} + \frac{11}{8 \left(x - 3\right)}

    The option matching the constants found is the correct answer.

Answer
38(x+1)−34(x−1)+118(x−3)\frac{3}{8 \left(x + 1\right)} - \frac{3}{4 \left(x - 1\right)} + \frac{11}{8 \left(x - 3\right)}

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