Hard A-Level Integration by parts Questions

Challenging, exam-style A-Level Integration by parts questions with worked solutions. Stretch yourself on the hardest integration by parts, LIATE, loop integral, area under curve problems.

integration by partsLIATEloop integralarea under curve
A-Level34 questionsStep-by-step solutions
Question 1
8 markschallenging
Using integration by parts, what is the exact value of ∫01xex dx\int_{0}^{1} x e^{x}\,dx?
Show worked solution

Worked solution

  1. Write down the integral

    ∫xex dx\int x e^{x}\,dx

    The integrand is a product of two different kinds of function, so integration by parts is appropriate.

  2. Recall the integration by parts formula

    ∫u dvdx dx=uv−∫v dudx dx\int u\,\frac{dv}{dx}\,dx=uv-\int v\,\frac{du}{dx}\,dx

    This rewrites the integral of a product in terms of a simpler integral.

  3. Choose uu using LIATE

    u=xu=x

    LIATE (logs, inverse-trig, algebraic, trig, exponential) selects the factor to differentiate; it should become simpler.

  4. Choose dvdx\dfrac{dv}{dx} (the other factor)

    dvdx=ex\frac{dv}{dx}=e^{x}

    The remaining factor is the part we integrate.

  5. Differentiate uu to find dudx\dfrac{du}{dx}

    dudx=1\frac{du}{dx}=1

    Differentiate the chosen uu.

  6. Integrate dvdx\dfrac{dv}{dx} to find vv

    v=exv=e^{x}

    Integrate the other factor; no constant of integration is needed at this stage.

  7. Collect the four ingredients

    u=x,dudx=1,dvdx=ex,v=exu=x,\quad \frac{du}{dx}=1,\quad \frac{dv}{dx}=e^{x},\quad v=e^{x}

    Having every piece ready makes the substitution reliable.

  8. Substitute into the formula

    ∫xex dx=xex−∫ex dx\int x e^{x}\,dx=x e^{x}-\int e^{x}\,dx

    Replace uvuv and ∫v dudx dx\int v\,\dfrac{du}{dx}\,dx using the pieces above.

  9. Look at the remaining integral

    ∫ex dx\int e^{x}\,dx

    This integral is simpler than the original.

  10. Evaluate the remaining integral

    ∫ex dx=ex\int e^{x}\,dx=e^{x}

    Integrate again — a second application of integration by parts may be required.

  11. Combine the two parts

    xex−(ex)x e^{x}-\left(e^{x}\right)

    Subtract the value of the remaining integral from uvuv.

  12. Simplify

    (x−1)ex\left(x - 1\right) e^{x}

    Collect the terms into a single expression.

  13. Add the constant of integration

    (x−1)ex+c\left(x - 1\right) e^{x}+c

    An indefinite integral must include the arbitrary constant +c+c.

  14. Check by differentiating

    ddx((x−1)ex)=xex\frac{d}{dx}\left(\left(x - 1\right) e^{x}\right)=x e^{x}

    Differentiating the answer returns the original integrand, confirming the result.

  15. Select the correct formula

    ∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du

    This is the standard statement of integration by parts.

Answer
11
Question 2
8 markschallenging
Which of the following is ∫x2e2x dx\int x^{2} e^{2 x}\,dx?
Show worked solution

Worked solution

  1. Write down the integral

    ∫x2e2x dx\int x^{2} e^{2 x}\,dx

    The integrand is a product of two different kinds of function, so integration by parts is appropriate.

  2. Recall the integration by parts formula

    ∫u dvdx dx=uv−∫v dudx dx\int u\,\frac{dv}{dx}\,dx=uv-\int v\,\frac{du}{dx}\,dx

    This rewrites the integral of a product in terms of a simpler integral.

  3. Choose uu using LIATE

    u=x2u=x^{2}

    LIATE (logs, inverse-trig, algebraic, trig, exponential) selects the factor to differentiate; it should become simpler.

  4. Choose dvdx\dfrac{dv}{dx} (the other factor)

    dvdx=e2x\frac{dv}{dx}=e^{2 x}

    The remaining factor is the part we integrate.

  5. Differentiate uu to find dudx\dfrac{du}{dx}

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

    Differentiate the chosen uu.

  6. Integrate dvdx\dfrac{dv}{dx} to find vv

    v=e2x2v=\frac{e^{2 x}}{2}

    Integrate the other factor; no constant of integration is needed at this stage.

  7. Collect the four ingredients

    u=x2,dudx=2x,dvdx=e2x,v=e2x2u=x^{2},\quad \frac{du}{dx}=2 x,\quad \frac{dv}{dx}=e^{2 x},\quad v=\frac{e^{2 x}}{2}

    Having every piece ready makes the substitution reliable.

  8. Substitute into the formula

    ∫x2e2x dx=x2e2x2−∫xe2x dx\int x^{2} e^{2 x}\,dx=\frac{x^{2} e^{2 x}}{2}-\int x e^{2 x}\,dx

    Replace uvuv and ∫v dudx dx\int v\,\dfrac{du}{dx}\,dx using the pieces above.

  9. Look at the remaining integral

    ∫xe2x dx\int x e^{2 x}\,dx

    This integral is simpler than the original.

  10. Evaluate the remaining integral

    ∫xe2x dx=(2x−1)e2x4\int x e^{2 x}\,dx=\frac{\left(2 x - 1\right) e^{2 x}}{4}

    Integrate again — a second application of integration by parts may be required.

  11. Combine the two parts

    x2e2x2−((2x−1)e2x4)\frac{x^{2} e^{2 x}}{2}-\left(\frac{\left(2 x - 1\right) e^{2 x}}{4}\right)

    Subtract the value of the remaining integral from uvuv.

  12. Simplify

    (2x2−2x+1)e2x4\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}

    Collect the terms into a single expression.

  13. Add the constant of integration

    (2x2−2x+1)e2x4+c\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}+c

    An indefinite integral must include the arbitrary constant +c+c.

  14. Check by differentiating

    ddx((2x2−2x+1)e2x4)=x2e2x\frac{d}{dx}\left(\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}\right)=x^{2} e^{2 x}

    Differentiating the answer returns the original integrand, confirming the result.

  15. Select the correct result

    ∫x2e2x dx=(2x2−2x+1)e2x4\int x^{2} e^{2 x}\,dx=\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}

    Integration by parts gives this antiderivative (plus a constant).

Answer
(2x2−2x+1)e2x4+c\frac{\left(2 x^{2} - 2 x + 1\right) e^{2 x}}{4}+c
Question 3
8 markschallenging
What is the correct first step when evaluating ∫x3ex dx\int x^{3} e^{x}\,dx?
Show worked solution

Worked solution

  1. Write down the integral

    ∫x3ex dx\int x^{3} e^{x}\,dx

    The integrand is a product of two different kinds of function, so integration by parts is appropriate.

  2. Recall the integration by parts formula

    ∫u dvdx dx=uv−∫v dudx dx\int u\,\frac{dv}{dx}\,dx=uv-\int v\,\frac{du}{dx}\,dx

    This rewrites the integral of a product in terms of a simpler integral.

  3. Choose uu using LIATE

    u=x3u=x^{3}

    LIATE (logs, inverse-trig, algebraic, trig, exponential) selects the factor to differentiate; it should become simpler.

  4. Choose dvdx\dfrac{dv}{dx} (the other factor)

    dvdx=ex\frac{dv}{dx}=e^{x}

    The remaining factor is the part we integrate.

  5. Differentiate uu to find dudx\dfrac{du}{dx}

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

    Differentiate the chosen uu.

  6. Integrate dvdx\dfrac{dv}{dx} to find vv

    v=exv=e^{x}

    Integrate the other factor; no constant of integration is needed at this stage.

  7. Collect the four ingredients

    u=x3,dudx=3x2,dvdx=ex,v=exu=x^{3},\quad \frac{du}{dx}=3 x^{2},\quad \frac{dv}{dx}=e^{x},\quad v=e^{x}

    Having every piece ready makes the substitution reliable.

  8. Substitute into the formula

    ∫x3ex dx=x3ex−∫3x2ex dx\int x^{3} e^{x}\,dx=x^{3} e^{x}-\int 3 x^{2} e^{x}\,dx

    Replace uvuv and ∫v dudx dx\int v\,\dfrac{du}{dx}\,dx using the pieces above.

  9. Look at the remaining integral

    ∫3x2ex dx\int 3 x^{2} e^{x}\,dx

    This integral is simpler than the original.

  10. Evaluate the remaining integral

    ∫3x2ex dx=(3x2−6x+6)ex\int 3 x^{2} e^{x}\,dx=\left(3 x^{2} - 6 x + 6\right) e^{x}

    Integrate again — a second application of integration by parts may be required.

  11. Combine the two parts

    x3ex−((3x2−6x+6)ex)x^{3} e^{x}-\left(\left(3 x^{2} - 6 x + 6\right) e^{x}\right)

    Subtract the value of the remaining integral from uvuv.

  12. Simplify

    (x3−3x2+6x−6)ex\left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}

    Collect the terms into a single expression.

  13. Add the constant of integration

    (x3−3x2+6x−6)ex+c\left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}+c

    An indefinite integral must include the arbitrary constant +c+c.

  14. Check by differentiating

    ddx((x3−3x2+6x−6)ex)=x3ex\frac{d}{dx}\left(\left(x^{3} - 3 x^{2} + 6 x - 6\right) e^{x}\right)=x^{3} e^{x}

    Differentiating the answer returns the original integrand, confirming the result.

  15. State the correct first step

    Split into u and dvdx, then apply parts.\text{Split into }u\text{ and }\frac{dv}{dx}\text{, then apply parts.}

    Identifying uu and dvdx\dfrac{dv}{dx} sets up the by-parts formula.

Answer
Write the integrand as udvdxu\dfrac{dv}{dx} and apply integration by parts
Question 4
8 markschallenging
Which of the following is ∫excos⁡(x) dx\int e^{x} \cos{\left(x \right)}\,dx?
Show worked solution

Worked solution

  1. Write down the integral

    ∫excos⁡(x) dx\int e^{x} \cos{\left(x \right)}\,dx

    The integrand is a product of two different kinds of function, so integration by parts is appropriate.

  2. Recall the integration by parts formula

    ∫u dvdx dx=uv−∫v dudx dx\int u\,\frac{dv}{dx}\,dx=uv-\int v\,\frac{du}{dx}\,dx

    This rewrites the integral of a product in terms of a simpler integral.

  3. Choose uu using LIATE

    u=cos⁡(x)u=\cos{\left(x \right)}

    LIATE (logs, inverse-trig, algebraic, trig, exponential) selects the factor to differentiate; it should become simpler.

  4. Choose dvdx\dfrac{dv}{dx} (the other factor)

    dvdx=ex\frac{dv}{dx}=e^{x}

    The remaining factor is the part we integrate.

  5. Differentiate uu to find dudx\dfrac{du}{dx}

    dudx=−sin⁡(x)\frac{du}{dx}=- \sin{\left(x \right)}

    Differentiate the chosen uu.

  6. Integrate dvdx\dfrac{dv}{dx} to find vv

    v=exv=e^{x}

    Integrate the other factor; no constant of integration is needed at this stage.

  7. Collect the four ingredients

    u=cos⁡(x),dudx=−sin⁡(x),dvdx=ex,v=exu=\cos{\left(x \right)},\quad \frac{du}{dx}=- \sin{\left(x \right)},\quad \frac{dv}{dx}=e^{x},\quad v=e^{x}

    Having every piece ready makes the substitution reliable.

  8. Substitute into the formula

    ∫excos⁡(x) dx=excos⁡(x)−∫−exsin⁡(x) dx\int e^{x} \cos{\left(x \right)}\,dx=e^{x} \cos{\left(x \right)}-\int - e^{x} \sin{\left(x \right)}\,dx

    Replace uvuv and ∫v dudx dx\int v\,\dfrac{du}{dx}\,dx using the pieces above.

  9. Look at the remaining integral

    ∫−exsin⁡(x) dx\int - e^{x} \sin{\left(x \right)}\,dx

    This integral is simpler than the original.

  10. Evaluate the remaining integral

    ∫−exsin⁡(x) dx=−exsin⁡(x)2+excos⁡(x)2\int - e^{x} \sin{\left(x \right)}\,dx=- \frac{e^{x} \sin{\left(x \right)}}{2} + \frac{e^{x} \cos{\left(x \right)}}{2}

    Integrate again — a second application of integration by parts may be required.

  11. Combine the two parts

    excos⁡(x)−(−exsin⁡(x)2+excos⁡(x)2)e^{x} \cos{\left(x \right)}-\left(- \frac{e^{x} \sin{\left(x \right)}}{2} + \frac{e^{x} \cos{\left(x \right)}}{2}\right)

    Subtract the value of the remaining integral from uvuv.

  12. Simplify

    2exsin⁡(x+π4)2\frac{\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}

    Collect the terms into a single expression.

  13. Add the constant of integration

    2exsin⁡(x+π4)2+c\frac{\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}+c

    An indefinite integral must include the arbitrary constant +c+c.

  14. Check by differentiating

    ddx(2exsin⁡(x+π4)2)=excos⁡(x)\frac{d}{dx}\left(\frac{\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}\right)=e^{x} \cos{\left(x \right)}

    Differentiating the answer returns the original integrand, confirming the result.

  15. Select the correct result

    ∫excos⁡(x) dx=2exsin⁡(x+π4)2\int e^{x} \cos{\left(x \right)}\,dx=\frac{\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}

    Integration by parts gives this antiderivative (plus a constant).

Answer
2exsin⁡(x+π4)2+c\frac{\sqrt{2} e^{x} \sin{\left(x + \frac{\pi}{4} \right)}}{2}+c
Question 5
8 markschallenging
To evaluate ∫x3ln⁡(x) dx\int x^{3} \ln{\left(x \right)}\,dx using integration by parts, which is the best choice of uu?
Show worked solution

Worked solution

  1. Write down the integral

    ∫x3ln⁡(x) dx\int x^{3} \ln{\left(x \right)}\,dx

    The integrand is a product of two different kinds of function, so integration by parts is appropriate.

  2. Recall the integration by parts formula

    ∫u dvdx dx=uv−∫v dudx dx\int u\,\frac{dv}{dx}\,dx=uv-\int v\,\frac{du}{dx}\,dx

    This rewrites the integral of a product in terms of a simpler integral.

  3. Choose uu using LIATE

    u=ln⁡(x)u=\ln{\left(x \right)}

    LIATE (logs, inverse-trig, algebraic, trig, exponential) selects the factor to differentiate; it should become simpler.

  4. Choose dvdx\dfrac{dv}{dx} (the other factor)

    dvdx=x3\frac{dv}{dx}=x^{3}

    The remaining factor is the part we integrate.

  5. Differentiate uu to find dudx\dfrac{du}{dx}

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

    Differentiate the chosen uu.

  6. Integrate dvdx\dfrac{dv}{dx} to find vv

    v=x44v=\frac{x^{4}}{4}

    Integrate the other factor; no constant of integration is needed at this stage.

  7. Collect the four ingredients

    u=ln⁡(x),dudx=1x,dvdx=x3,v=x44u=\ln{\left(x \right)},\quad \frac{du}{dx}=\frac{1}{x},\quad \frac{dv}{dx}=x^{3},\quad v=\frac{x^{4}}{4}

    Having every piece ready makes the substitution reliable.

  8. Substitute into the formula

    ∫x3ln⁡(x) dx=x4ln⁡(x)4−∫x34 dx\int x^{3} \ln{\left(x \right)}\,dx=\frac{x^{4} \ln{\left(x \right)}}{4}-\int \frac{x^{3}}{4}\,dx

    Replace uvuv and ∫v dudx dx\int v\,\dfrac{du}{dx}\,dx using the pieces above.

  9. Look at the remaining integral

    ∫x34 dx\int \frac{x^{3}}{4}\,dx

    This integral is simpler than the original.

  10. Evaluate the remaining integral

    ∫x34 dx=x416\int \frac{x^{3}}{4}\,dx=\frac{x^{4}}{16}

    Integrate again — a second application of integration by parts may be required.

  11. Combine the two parts

    x4ln⁡(x)4−(x416)\frac{x^{4} \ln{\left(x \right)}}{4}-\left(\frac{x^{4}}{16}\right)

    Subtract the value of the remaining integral from uvuv.

  12. Simplify

    x4(4ln⁡(x)−1)16\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}

    Collect the terms into a single expression.

  13. Add the constant of integration

    x4(4ln⁡(x)−1)16+c\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}+c

    An indefinite integral must include the arbitrary constant +c+c.

  14. Check by differentiating

    ddx(x4(4ln⁡(x)−1)16)=x3ln⁡(x)\frac{d}{dx}\left(\frac{x^{4} \left(4 \ln{\left(x \right)} - 1\right)}{16}\right)=x^{3} \ln{\left(x \right)}

    Differentiating the answer returns the original integrand, confirming the result.

  15. Select the best choice of uu

    u=ln⁡(x)u=\ln{\left(x \right)}

    By LIATE this factor is differentiated because it becomes simpler.

Answer
u=ln⁡(x)u=\ln{\left(x \right)}

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