Hard A-Level Integration by substitution Questions

Challenging, exam-style A-Level Integration by substitution questions with worked solutions. Stretch yourself on the hardest substitution, reverse chain rule, logarithmic, exponential problems.

substitutionreverse chain rulelogarithmicexponentialsurd
A-Level34 questionsStep-by-step solutions
Question 1
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.
Show worked solution

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
Question 2
8 markschallenging
Explain which substitution reduces ∫013x2x3+1 dx\int_{0}^{1} \frac{3 x^{2}}{x^{3} + 1}\,dx to a standard integral.
Show worked solution

Worked solution

  1. Write down the definite integral

    ∫013x2x3+1 dx\int_{0}^{1} \frac{3 x^{2}}{x^{3} + 1}\,dx

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

  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 integrand.

  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. Make dx the subject

    dx=du3x2dx=\frac{du}{3 x^{2}}

    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=2x=1\ \Rightarrow\ u=2

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

  9. Rewrite the integral in terms of u

    ∫121u du\int_{1}^{2} \frac{1}{u}\,du

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

  10. Integrate with respect to u

    [ln⁡(u)]12\left[\ln{\left(u \right)}\right]_{1}^{2}

    Integrate and prepare to evaluate between the new limits.

  11. Evaluate at the upper limit

    ln⁡(2)\ln{\left(2 \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⁡(2)−(0)=ln⁡(2)\ln{\left(2 \right)}-\left(0\right)=\ln{\left(2 \right)}

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

  14. Simplify the exact value

    ln⁡(2)\ln{\left(2 \right)}

    Simplify to obtain the exact value of the integral.

  15. 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 3
8 markschallenging
For the integral ∫01xex2 dx\int_{0}^{1} x e^{x^{2}}\,dx, state the most suitable substitution.
Show worked solution

Worked solution

  1. Write down the definite integral

    ∫01xex2 dx\int_{0}^{1} x e^{x^{2}}\,dx

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

  2. Identify the inner function

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

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

  3. Choose the substitution

    u=x2u=x^{2}

    Replacing the inner function by u simplifies the integrand.

  4. Differentiate the substitution

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

    Differentiate u with respect to x.

  5. Express du in terms of dx

    du=2x dxdu=2 x\,dx

    Multiply both sides by dx to prepare for the substitution.

  6. Make dx the subject

    dx=du2xdx=\frac{du}{2 x}

    Rearranging lets us replace dx wherever it appears.

  7. Change the lower limit

    x=0 ⇒ u=0x=0\ \Rightarrow\ u=0

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

  8. Change the upper limit

    x=1 ⇒ u=1x=1\ \Rightarrow\ u=1

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

  9. Rewrite the integral in terms of u

    ∫01eu2 du\int_{0}^{1} \frac{e^{u}}{2}\,du

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

  10. Integrate with respect to u

    [eu2]01\left[\frac{e^{u}}{2}\right]_{0}^{1}

    Integrate and prepare to evaluate between the new limits.

  11. Evaluate at the upper limit

    e2\frac{e}{2}

    Substitute the upper u-limit into the antiderivative.

  12. Evaluate at the lower limit

    12\frac{1}{2}

    Substitute the lower u-limit into the antiderivative.

  13. Subtract to find the value

    e2−(12)=−12+e2\frac{e}{2}-\left(\frac{1}{2}\right)=- \frac{1}{2} + \frac{e}{2}

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

  14. Simplify the exact value

    −12+e2- \frac{1}{2} + \frac{e}{2}

    Simplify to obtain the exact value of the integral.

  15. Select the correct option

    u=x2u=x^{2}

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

Answer
u=x2u=x^{2}
Question 4
8 markschallenging
Which of the following is the value of ∫01xx2+1 dx\int_{0}^{1} x \sqrt{x^{2} + 1}\,dx?
Show worked solution

Worked solution

  1. Write down the definite integral

    ∫01xx2+1 dx\int_{0}^{1} x \sqrt{x^{2} + 1}\,dx

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

  2. Identify the inner function

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

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

  3. Choose the substitution

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

    Replacing the inner function by u simplifies the integrand.

  4. Differentiate the substitution

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

    Differentiate u with respect to x.

  5. Express du in terms of dx

    du=2x dxdu=2 x\,dx

    Multiply both sides by dx to prepare for the substitution.

  6. Make dx the subject

    dx=du2xdx=\frac{du}{2 x}

    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=2x=1\ \Rightarrow\ u=2

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

  9. Rewrite the integral in terms of u

    ∫12u2 du\int_{1}^{2} \frac{\sqrt{u}}{2}\,du

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

  10. Integrate with respect to u

    [u323]12\left[\frac{u^{\frac{3}{2}}}{3}\right]_{1}^{2}

    Integrate and prepare to evaluate between the new limits.

  11. Evaluate at the upper limit

    223\frac{2 \sqrt{2}}{3}

    Substitute the upper u-limit into the antiderivative.

  12. Evaluate at the lower limit

    13\frac{1}{3}

    Substitute the lower u-limit into the antiderivative.

  13. Subtract to find the value

    223−(13)=−13+223\frac{2 \sqrt{2}}{3}-\left(\frac{1}{3}\right)=- \frac{1}{3} + \frac{2 \sqrt{2}}{3}

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

  14. Simplify the exact value

    −13+223- \frac{1}{3} + \frac{2 \sqrt{2}}{3}

    Simplify to obtain the exact value of the integral.

  15. Select the correct option

    ∫01xx2+1 dx=−13+223\int_{0}^{1} x \sqrt{x^{2} + 1}\,dx=- \frac{1}{3} + \frac{2 \sqrt{2}}{3}

    This matches the value found by changing the limits.

Answer
−13+223- \frac{1}{3} + \frac{2 \sqrt{2}}{3}
Question 5
8 markschallenging
Which of the following is the value of ∫01xx2+9 dx\int_{0}^{1} \frac{x}{\sqrt{x^{2} + 9}}\,dx?
Show worked solution

Worked solution

  1. Write down the definite integral

    ∫01xx2+9 dx\int_{0}^{1} \frac{x}{\sqrt{x^{2} + 9}}\,dx

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

  2. Identify the inner function

    g(x)=x2+9,g′(x)=2xg(x)=x^{2} + 9,\quad g'(x)=2 x

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

  3. Choose the substitution

    u=x2+9u=x^{2} + 9

    Replacing the inner function by u simplifies the integrand.

  4. Differentiate the substitution

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

    Differentiate u with respect to x.

  5. Express du in terms of dx

    du=2x dxdu=2 x\,dx

    Multiply both sides by dx to prepare for the substitution.

  6. Make dx the subject

    dx=du2xdx=\frac{du}{2 x}

    Rearranging lets us replace dx wherever it appears.

  7. Change the lower limit

    x=0 ⇒ u=9x=0\ \Rightarrow\ u=9

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

  8. Change the upper limit

    x=1 ⇒ u=10x=1\ \Rightarrow\ u=10

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

  9. Rewrite the integral in terms of u

    ∫91012u du\int_{9}^{10} \frac{1}{2 \sqrt{u}}\,du

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

  10. Integrate with respect to u

    [u]910\left[\sqrt{u}\right]_{9}^{10}

    Integrate and prepare to evaluate between the new limits.

  11. Evaluate at the upper limit

    10\sqrt{10}

    Substitute the upper u-limit into the antiderivative.

  12. Evaluate at the lower limit

    33

    Substitute the lower u-limit into the antiderivative.

  13. Subtract to find the value

    10−(3)=−3+10\sqrt{10}-\left(3\right)=-3 + \sqrt{10}

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

  14. Simplify the exact value

    −3+10-3 + \sqrt{10}

    Simplify to obtain the exact value of the integral.

  15. Select the correct option

    ∫01xx2+9 dx=−3+10\int_{0}^{1} \frac{x}{\sqrt{x^{2} + 9}}\,dx=-3 + \sqrt{10}

    This matches the value found by changing the limits.

Answer
−3+10-3 + \sqrt{10}

Unlock 29 more Integration by substitution questions

Create a free account to work through every A-Level Integration by substitution question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Integration by substitution practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Areas and further applicationsMathematical argument and notationArithmetic sequences and seriesBinomial expansion (rational powers)Binomial expansionCircles