A-Level Integration by substitution Practice Questions

Free A-Level Integration by substitution practice questions with full step-by-step worked solutions. Covers substitution, reverse chain rule, exponential, logarithmic. Practise exam-style problems and check your method.

substitutionreverse chain ruleexponentiallogarithmicsurdtrigonometric
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
Find ∫2x(x2+1)5 dx\int 2 x \left(x^{2} + 1\right)^{5}\,dx.
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Worked solution

  1. Choose the substitution and differentiate

    u=x2+1,du=2x dxu=x^{2} + 1,\quad du=2 x\,dx

    The derivative of the inner function is a factor of the integrand.

  2. Rewrite in terms of u and integrate

    ∫u5 du=u66+c\int u^{5}\,du=\frac{u^{6}}{6}+c

    Integrate the standard expression in u.

  3. State the final answer

    ∫2x(x2+1)5 dx=(x2+1)66+c\int 2 x \left(x^{2} + 1\right)^{5}\,dx=\frac{\left(x^{2} + 1\right)^{6}}{6}+c

    Remember to include the constant of integration.

Answer
(x2+1)66+c\frac{\left(x^{2} + 1\right)^{6}}{6}+c
Question 2
2 markseasy
Describe the substitution needed to evaluate ∫6x(3x2−2)4 dx\int 6 x \left(3 x^{2} - 2\right)^{4}\,dx.
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Worked solution

  1. Look for an inner function whose derivative appears

    ∫6x(3x2−2)4 dx\int 6 x \left(3 x^{2} - 2\right)^{4}\,dx

    A substitution works when g'(x) is a factor of the integrand.

  2. Check the derivative of the inner function

    ddx(3x2−2)=6x\frac{d}{dx}\left(3 x^{2} - 2\right)=6 x

    This derivative matches the remaining factor, so the substitution is valid.

  3. Select the correct option

    u=3x2−2u=3 x^{2} - 2

    The inner function's derivative is a factor of the integrand.

Answer
u=3x2−2u=3 x^{2} - 2
Question 3
3 marksintermediate
Describe the substitution needed to evaluate ∫exex+1 dx\int \frac{e^{x}}{e^{x} + 1}\,dx.
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Worked solution

  1. Write down the integral

    ∫exex+1 dx\int \frac{e^{x}}{e^{x} + 1}\,dx

    We evaluate this using a substitution to reverse the chain rule.

  2. Choose the substitution

    u=ex+1u=e^{x} + 1

    Replacing the inner function by u simplifies the integral.

  3. Differentiate the substitution

    dudx=ex\frac{du}{dx}=e^{x}

    Differentiate u with respect to x.

  4. Rewrite the integral in terms of u

    ∫1u du\int \frac{1}{u}\,du

    Every x has now been replaced, leaving a standard integral in u.

  5. Integrate with respect to u

    ∫1u du=ln⁡(u)+c\int \frac{1}{u}\,du=\ln{\left(u \right)}+c

    Integrate the simplified expression, adding a constant of integration.

  6. Select the correct option

    u=ex+1u=e^{x} + 1

    The inner function's derivative is a factor of the integrand.

Answer
u=ex+1u=e^{x} + 1
Question 4
5 markshard
Describe the substitution needed to evaluate ∫x2(x3+1)4 dx\int x^{2} \left(x^{3} + 1\right)^{4}\,dx.
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Worked solution

  1. Write down the integral

    ∫x2(x3+1)4 dx\int x^{2} \left(x^{3} + 1\right)^{4}\,dx

    We evaluate this using a substitution to reverse the chain rule.

  2. Identify the inner function

    g(x)=x3+1,g′(x)=3x2g(x)=x^{3} + 1,\quad g'(x)=3 x^{2}

    Its derivative is a factor of the integrand, which signals a substitution.

  3. Choose the substitution

    u=x3+1u=x^{3} + 1

    Replacing the inner function by u simplifies the integral.

  4. Differentiate the substitution

    dudx=3x2\frac{du}{dx}=3 x^{2}

    Differentiate u with respect to x.

  5. Express du in terms of dx

    du=3x2 dxdu=3 x^{2}\,dx

    Multiply both sides by dx to prepare for the substitution.

  6. Rewrite the integral in terms of u

    ∫u43 du\int \frac{u^{4}}{3}\,du

    Every x has now been replaced, leaving a standard integral in u.

  7. Recall the relevant standard integral

    ∫un du=un+1n+1+c(n≠−1)\int u^{n}\,du=\frac{u^{n+1}}{n+1}+c\quad(n\neq-1)

    This is the standard result we now apply.

  8. Integrate with respect to u

    ∫u43 du=u515+c\int \frac{u^{4}}{3}\,du=\frac{u^{5}}{15}+c

    Integrate the simplified expression, adding a constant of integration.

  9. Return to the original variable

    (x3+1)515+c\frac{\left(x^{3} + 1\right)^{5}}{15}+c

    Back-substitute u=x^{3} + 1 so the answer is in terms of x.

  10. Select the correct option

    u=x3+1u=x^{3} + 1

    The inner function's derivative is a factor of the integrand.

Answer
u=x3+1u=x^{3} + 1
Question 5
8 markschallenging
Describe the substitution needed to evaluate ∫012x+1x2+x+1 dx\int_{0}^{1} \frac{2 x + 1}{x^{2} + x + 1}\,dx.
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Worked solution

  1. Write down the definite integral

    ∫012x+1x2+x+1 dx\int_{0}^{1} \frac{2 x + 1}{x^{2} + x + 1}\,dx

    We evaluate this with a substitution and will change the limits.

  2. Identify the inner function

    g(x)=x2+x+1,g′(x)=2x+1g(x)=x^{2} + x + 1,\quad g'(x)=2 x + 1

    Its derivative is a factor of the integrand, which signals a substitution.

  3. Choose the substitution

    u=x2+x+1u=x^{2} + x + 1

    Replacing the inner function by u simplifies the integrand.

  4. Differentiate the substitution

    dudx=2x+1\frac{du}{dx}=2 x + 1

    Differentiate u with respect to x.

  5. Express du in terms of dx

    du=2x+1 dxdu=2 x + 1\,dx

    Multiply both sides by dx to prepare for the substitution.

  6. Make dx the subject

    dx=du2x+1dx=\frac{du}{2 x + 1}

    Rearranging lets us replace dx wherever it appears.

  7. Change the lower limit

    x=0 ⇒ u=1x=0\ \Rightarrow\ u=1

    Substitute the lower x-limit into u=g(x) to get the new lower limit.

  8. Change the upper limit

    x=1 ⇒ u=3x=1\ \Rightarrow\ u=3

    Substitute the upper x-limit into u=g(x) to get the new upper limit.

  9. Rewrite the integral in terms of u

    ∫131u du\int_{1}^{3} \frac{1}{u}\,du

    With the limits changed, no back-substitution is needed.

  10. Integrate with respect to u

    [ln⁡(u)]13\left[\ln{\left(u \right)}\right]_{1}^{3}

    Integrate and prepare to evaluate between the new limits.

  11. Evaluate at the upper limit

    ln⁡(3)\ln{\left(3 \right)}

    Substitute the upper u-limit into the antiderivative.

  12. Evaluate at the lower limit

    00

    Substitute the lower u-limit into the antiderivative.

  13. Subtract to find the value

    ln⁡(3)−(0)=ln⁡(3)\ln{\left(3 \right)}-\left(0\right)=\ln{\left(3 \right)}

    The definite integral is the upper value minus the lower value.

  14. Simplify the exact value

    ln⁡(3)\ln{\left(3 \right)}

    Simplify to obtain the exact value of the integral.

  15. Select the correct option

    u=x2+x+1u=x^{2} + x + 1

    The inner function's derivative is a factor of the integrand.

Answer
u=x2+x+1u=x^{2} + x + 1

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