A-Level Integration with partial fractions Practice Questions

Free A-Level Integration with partial fractions practice questions with full step-by-step worked solutions. Covers partial fractions, integration. Practise exam-style problems and check your method.

partial fractionsintegration
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Find ∫5x−2 dx\int \frac{5}{x - 2}\,dx.
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Worked solution

  1. Write down the integral

    ∫5x−2 dx\int \frac{5}{x - 2}\,dx

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

  2. Integrate using the standard logarithm result

    ∫5x−2 dx=5ln⁡(x−2)\int \frac{5}{x - 2}\,dx=5 \ln{\left(x - 2 \right)}

    Recall \int \frac{k}{x-a}\,dx=k\ln|x-a|+c.

  3. Add the constant of integration

    5ln⁡(x−2)+c5 \ln{\left(x - 2 \right)} + c

    This is an indefinite integral, so we include +c.

Answer
5ln⁡(x−2)+c5 \ln{\left(x - 2 \right)} + c
Question 2
2 markseasy
Which of these is the correct partial fraction form for 2x+1(x−4)(x+1)\frac{2 x + 1}{\left(x - 4\right) \left(x + 1\right)}?
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Worked solution

  1. Factorise the denominator

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

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

  2. Set up the partial fraction form

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

    Write one constant over each distinct linear factor.

  3. Select the correct option

    Ax+1+Bx−4\frac{A}{x + 1}+\frac{B}{x - 4}

    The option matching one constant over each linear factor is the correct answer.

Answer
Ax+1+Bx−4\frac{A}{x + 1}+\frac{B}{x - 4}
Question 3
3 marksintermediate
Which of these is the correct partial fraction form for 6x−1(x−5)(x−2)\frac{6 x - 1}{\left(x - 5\right) \left(x - 2\right)}?
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Worked solution

  1. Factorise the denominator

    x2−7x+10=(x−5)(x−2)x^{2} - 7 x + 10=\left(x - 5\right) \left(x - 2\right)

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

  2. Set up the partial fraction form

    6x−1(x−5)(x−2)≡Ax−2+Bx−5\frac{6 x - 1}{\left(x - 5\right) \left(x - 2\right)}\equiv \frac{A}{x - 2}+\frac{B}{x - 5}

    Write one constant over each distinct linear factor.

  3. Multiply through by the denominator

    6x−1≡A(x−5)+B(x−2)6 x - 1\equiv A\left(x - 5\right)+B\left(x - 2\right)

    Clearing the fractions gives an identity valid for all x.

  4. Substitute x=2 to find A

    A=−113A=- \frac{11}{3}

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

  5. Substitute x=5 to find B

    B=293B=\frac{29}{3}

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

  6. Select the correct option

    Ax−2+Bx−5\frac{A}{x - 2}+\frac{B}{x - 5}

    The option matching one constant over each linear factor is the correct answer.

Answer
Ax−2+Bx−5\frac{A}{x - 2}+\frac{B}{x - 5}
Question 4
5 markshard
When x+2(x−1)(x+5)\frac{x + 2}{\left(x - 1\right) \left(x + 5\right)} is written as Ax+5+Bx−1\frac{A}{x + 5}+\frac{B}{x - 1}, find the value of AA.
Show worked solution

Worked solution

  1. Write down the integral

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

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

  2. Factorise the denominator

    x2+4x−5=(x−1)(x+5)x^{2} + 4 x - 5=\left(x - 1\right) \left(x + 5\right)

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

  3. Set up the partial fraction form

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

    Write one constant over each distinct linear factor.

  4. Multiply through by the denominator

    x+2≡A(x−1)+B(x+5)x + 2\equiv A\left(x - 1\right)+B\left(x + 5\right)

    Clearing the fractions gives an identity valid for all x.

  5. Substitute x=-5 to find A

    A=12A=\frac{1}{2}

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

  6. Substitute x=1 to find B

    B=12B=\frac{1}{2}

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

  7. Collect the constants

    A=12, B=12A=\frac{1}{2},\ B=\frac{1}{2}

    These constants complete the decomposition.

  8. Write the partial fraction decomposition

    x+2(x−1)(x+5)≡12(x+5)+12(x−1)\frac{x + 2}{\left(x - 1\right) \left(x + 5\right)}\equiv \frac{1}{2 \left(x + 5\right)} + \frac{1}{2 \left(x - 1\right)}

    Each simple fraction is now easy to integrate.

  9. Integrate the decomposition

    ∫x+2(x−1)(x+5) dx=ln⁡(x2+4x−5)2\int \frac{x + 2}{\left(x - 1\right) \left(x + 5\right)}\,dx=\frac{\ln{\left(x^{2} + 4 x - 5 \right)}}{2}

    Integrate each simple fraction to a logarithm.

  10. State the value of A

    A=12A=\frac{1}{2}

    This is the constant over the first linear factor.

Answer
A=12A=\frac{1}{2}
Question 5
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

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