Hard Further Maths Further calculus Questions

Challenging, exam-style Further Maths Further calculus questions with worked solutions. Stretch yourself on the hardest improper-integrals, partial-fractions, inverse-tangent-integral, inverse-sine-integral problems.

improper-integralspartial-fractionsinverse-tangent-integralinverse-sine-integralmean-valuepolynomial-integration
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
Which of the following is the correct partial fraction decomposition of 5x+3(x+1)(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}?
Show worked solution

Worked solution

  1. Write down the fraction to be decomposed

    5x+3(x+1)(x2+9)=5x+3(x+1)(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}=\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}

    The denominator is already a product of irreducible factors.

  2. Factorise the denominator

    x3+x2+9x+9=(x+1)(x2+9)x^{3}+x^{2}+9x+9=\left(x+1\right)\left(x^{2}+9\right)

    The denominator must be a product of irreducible factors before it can be split up.

  3. Write the integrand in partial fractions

    5x+3(x+1)(x2+9)≡A(x+1)+Bx+C(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}\equiv \frac{A}{\left(x+1\right)}+\frac{Bx+C}{\left(x^{2}+9\right)}

    Each factor contributes one term for every power up to its multiplicity; a quadratic factor carries a linear numerator.

  4. Multiply through by the denominator

    5x+3≡A(x2+9)+(Bx+C)(x+1)5x+3\equiv A\left(x^{2}+9\right)+\left(Bx+C\right)\left(x+1\right)

    Clearing the fractions leaves an identity that holds for every xx.

  5. Substitute x=−1x=-1 to find AA

    −2=A×(10)⇒A=−15-2=A\times\left(10\right)\quad\Rightarrow\quad A=-\frac{1}{5}

    Choosing the root of a factor kills every other term of the identity at once.

  6. Compare coefficients to find the remaining constants

    B=15,C=245B=\frac{1}{5},\quad C=\frac{24}{5}

    Equating the coefficients of each power of xx on both sides of the identity gives enough equations for the constants left over.

  7. Check the decomposition by recombining the fractions

    −15(x+1)+x+245(x2+9)≡5x+3(x+1)(x2+9)-\frac{1}{5\left(x+1\right)}+\frac{x+24}{5\left(x^{2}+9\right)}\equiv\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}

    Putting the partial fractions back over a common denominator must return the original fraction.

  8. Test the identity at x=2x=2

    LHS=13,RHS=13\text{LHS}=\frac{1}{3},\quad\text{RHS}=\frac{1}{3}

    Both sides of an identity must agree at every value of xx, so a single test value is a fast check on the constants.

  9. Recall the pp-test at infinity

    ∫1∞1xp dx converges  ⟺  p>1\int_{1}^{\infty}\frac{1}{x^{p}}\,dx\ \text{converges}\iff p>1

    The tail must decay strictly faster than 1x\frac{1}{x}.

  10. Recall the pp-test at a singularity

    ∫011xp dx converges  ⟺  p<1\int_{0}^{1}\frac{1}{x^{p}}\,dx\ \text{converges}\iff p<1

    Near the singularity the blow-up must be slower than 1x\frac{1}{x}.

  11. Note the limit of a decaying exponential

    lim⁡t→∞tme−kt=0(k>0)\lim_{t\to\infty}t^{m}e^{-kt}=0\quad\left(k>0\right)

    An exponential beats every power, so the product tends to zero.

  12. Note the limit of a negative power

    lim⁡t→∞1tm=0(m>0)\lim_{t\to\infty}\frac{1}{t^{m}}=0\quad\left(m>0\right)

    Any positive power of tt in a denominator drives the term to zero.

  13. Note the limit of the inverse tangent

    lim⁡t→∞arctan⁡t=π2\lim_{t\to\infty}\arctan t=\frac{\pi}{2}

    The graph of arctan⁡\arctan has a horizontal asymptote at π2\frac{\pi}{2}.

  14. Recall the logarithm law used to combine two logarithms

    ln⁡A−ln⁡B=ln⁡(AB)\ln A-\ln B=\ln\left(\frac{A}{B}\right)

    A difference of logarithms is the logarithm of a quotient.

  15. Recall the exact values of the inverse tangent

    arctan⁡1=π4,arctan⁡3=π3\arctan1=\frac{\pi}{4},\quad\arctan\sqrt{3}=\frac{\pi}{3}

    Exact values let the answer be given in terms of π\pi.

  16. Select the correct decomposition

    5x+3(x+1)(x2+9)≡−15(x+1)+x+245(x2+9)\frac{5x+3}{\left(x+1\right)\left(x^{2}+9\right)}\equiv -\frac{1}{5\left(x+1\right)}+\frac{x+24}{5\left(x^{2}+9\right)}

    This is the only option that is identically equal to the original fraction.

Answer
−15(x+1)+x+245(x2+9)-\frac{1}{5\left(x+1\right)}+\frac{x+24}{5\left(x^{2}+9\right)}
Question 2
9 markschallenging
Which of the following is ∫4x(x−1)(x2+1) dx\int \frac{4x}{\left(x-1\right)\left(x^{2}+1\right)}\,dx?
Show worked solution

Worked solution

  1. Write down the integral to be found

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

    There are no limits, so the answer is a family of functions.

  2. Factorise the denominator

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

    The denominator must be a product of irreducible factors before it can be split up.

  3. Write the integrand in partial fractions

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

    Each factor contributes one term for every power up to its multiplicity; a quadratic factor carries a linear numerator.

  4. Multiply through by the denominator

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

    Clearing the fractions leaves an identity that holds for every xx.

  5. Substitute x=1x=1 to find AA

    4=A×(2)⇒A=24=A\times\left(2\right)\quad\Rightarrow\quad A=2

    Choosing the root of a factor kills every other term of the identity at once.

  6. Compare coefficients to find the remaining constants

    B=−2,C=2B=-2,\quad C=2

    Equating the coefficients of each power of xx on both sides of the identity gives enough equations for the constants left over.

  7. State the partial fraction form

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

    This is the integrand written as a sum of standard fractions.

  8. Split the term over the quadratic factor

    px+qx2+a2=p2⋅2xx2+a2+qx2+a2\frac{px+q}{x^{2}+a^{2}}=\frac{p}{2}\cdot\frac{2x}{x^{2}+a^{2}}+\frac{q}{x^{2}+a^{2}}

    The first piece is a logarithm and the second is an inverse tangent.

  9. Check the antiderivative by differentiating it

    ddx(2ln⁡(x−1)−ln⁡(x2+1)+2arctan⁡(x))=4x(x−1)(x2+1)\frac{d}{dx}\left(2\ln{\left(x-1\right)}-\ln{\left(x^{2}+1\right)}+2\arctan{\left(x\right)}\right)=\frac{4x}{\left(x-1\right)\left(x^{2}+1\right)}

    Differentiating the answer must return the integrand exactly.

  10. Do not forget the constant of integration

    +c+c

    An indefinite integral is only determined up to an additive constant.

  11. Recall the integration by parts formula

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

    Integration by parts is the product rule read backwards.

  12. Recall the pp-test at infinity

    ∫1∞1xp dx converges  ⟺  p>1\int_{1}^{\infty}\frac{1}{x^{p}}\,dx\ \text{converges}\iff p>1

    The tail must decay strictly faster than 1x\frac{1}{x}.

  13. Recall the pp-test at a singularity

    ∫011xp dx converges  ⟺  p<1\int_{0}^{1}\frac{1}{x^{p}}\,dx\ \text{converges}\iff p<1

    Near the singularity the blow-up must be slower than 1x\frac{1}{x}.

  14. Note the limit of a decaying exponential

    lim⁡t→∞tme−kt=0(k>0)\lim_{t\to\infty}t^{m}e^{-kt}=0\quad\left(k>0\right)

    An exponential beats every power, so the product tends to zero.

  15. State the antiderivative

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

    This is the general antiderivative of the integrand.

Answer
2ln⁡(x−1)−ln⁡(x2+1)+2arctan⁡(x)+c2\ln{\left(x-1\right)}-\ln{\left(x^{2}+1\right)}+2\arctan{\left(x\right)}+c
Question 3
9 markschallenging
Which of the following is the value of ∫−3∞1x2+6x+13 dx\int_{-3}^{\infty}\frac{1}{x^{2}+6x+13}\,dx?
Show worked solution

Worked solution

  1. Write down the integral to be evaluated

    I=∫−3∞1x2+6x+13 dxI=\int_{-3}^{\infty}\frac{1}{x^{2}+6x+13}\,dx

    Identify the integrand and the limits before choosing a method.

  2. State why the integral is improper

    the interval is unbounded above\text{the interval is unbounded above}

    This integral is improper because the upper limit of integration is infinite.

  3. Replace the awkward limit by a variable and take a limit

    ∫−3∞1x2+6x+13 dx=lim⁡t→∞∫−3t1x2+6x+13 dx\int_{-3}^{\infty}\frac{1}{x^{2}+6x+13}\,dx=\lim_{t\to\infty}\int_{-3}^{t}\frac{1}{x^{2}+6x+13}\,dx

    The improper integral is DEFINED as this limit; replacing the offending limit by tt makes every step that follows a legitimate definite integral.

  4. Complete the square in the denominator

    x2+6x+13=(x+3)2+4x^{2}+6x+13=\left(x+3\right)^{2}+4

    The denominator is now in the standard form u2+a2u^{2}+a^{2} with u=x+3u=x+3.

  5. Note that the substitution is a translation

    u=x+3⇒du=dxu=x+3\quad\Rightarrow\quad du=dx

    A shift in xx leaves dxdx unchanged, so no extra factor appears.

  6. Match the denominator with the standard form

    a2=4⇒a=2a^{2}=4\quad\Rightarrow\quad a=2

    Reading off aa is all that is needed to quote the standard integral.

  7. Write down the antiderivative

    F(x)=arctan⁡(x2+32)2F(x)=\frac{\arctan{\left(\frac{x}{2}+\frac{3}{2}\right)}}{2}

    This is the standard inverse-tangent result with a=2a=2.

  8. Check the antiderivative by differentiating it

    ddx(arctan⁡(x2+32)2)=1x2+6x+13\frac{d}{dx}\left(\frac{\arctan{\left(\frac{x}{2}+\frac{3}{2}\right)}}{2}\right)=\frac{1}{x^{2}+6x+13}

    Differentiating FF must return the integrand exactly; if it does not, the integration is wrong.

  9. Apply the limits of integration

    ∫−3∞1x2+6x+13 dx=[arctan⁡(x2+32)2]−3∞\int_{-3}^{\infty}\frac{1}{x^{2}+6x+13}\,dx=\left[\frac{\arctan{\left(\frac{x}{2}+\frac{3}{2}\right)}}{2}\right]_{-3}^{\infty}

    The value is the antiderivative at the top minus the antiderivative at the bottom, with a limit taken at any improper endpoint.

  10. Take the limit of the antiderivative as t→∞t\to\infty

    lim⁡t→∞(arctan⁡(t2+32)2)=π4\lim_{t\to\infty}\left(\frac{\arctan{\left(\frac{t}{2}+\frac{3}{2}\right)}}{2}\right)=\frac{\pi}{4}

    This limit IS the improper integral; no substitution of ∞\infty is ever made.

  11. Evaluate the antiderivative at the lower limit

    F(−3)=0F\left(-3\right)=0

    The lower limit is an ordinary number, so it is simply substituted.

  12. Subtract to obtain the value of the integral

    ∫−3∞1x2+6x+13 dx=(π4)−(0)=π4\int_{-3}^{\infty}\frac{1}{x^{2}+6x+13}\,dx=\left(\frac{\pi}{4}\right)-\left(0\right)=\frac{\pi}{4}

    This is the exact value of the integral.

  13. Check the answer numerically

    I≈0.785398I\approx0.785398

    High-precision quadrature agrees with the exact value to six significant figures.

  14. Check that the answer is plausible

    1x2+6x+13>0 on the interval⇒I>0\frac{1}{x^{2}+6x+13}>0\ \text{on the interval}\Rightarrow I>0

    A positive integrand over an interval of positive width must give a positive answer.

  15. Recall the integral of an exponential

    ∫ekx dx=1kekx+c\int e^{kx}\,dx=\frac{1}{k}e^{kx}+c

    Dividing by the coefficient of xx in the index undoes the chain rule.

  16. Recall the integration by parts formula

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

    Integration by parts is the product rule read backwards.

  17. State the exact value

    I=π4I=\frac{\pi}{4}

    This is the exact answer required.

Answer
π4\frac{\pi}{4}
Question 4
9 markschallenging
Which of the following statements about the improper integral ∫0∞xe−2x dx\int_{0}^{\infty}xe^{-2x}\,dx is correct?
Show worked solution

Worked solution

  1. Write down the integral to be evaluated

    I=∫0∞xe−2x dxI=\int_{0}^{\infty}xe^{-2x}\,dx

    Identify the integrand and the limits before choosing a method.

  2. State why the integral is improper

    the interval is unbounded above\text{the interval is unbounded above}

    This integral is improper because the upper limit of integration is infinite.

  3. Replace the awkward limit by a variable and take a limit

    ∫0∞xe−2x dx=lim⁡t→∞∫0txe−2x dx\int_{0}^{\infty}xe^{-2x}\,dx=\lim_{t\to\infty}\int_{0}^{t}xe^{-2x}\,dx

    The improper integral is DEFINED as this limit; replacing the offending limit by tt makes every step that follows a legitimate definite integral.

  4. Set up integration by parts

    u=x,dvdx=e−2xu=x,\quad\frac{dv}{dx}=e^{-2x}

    The polynomial is chosen as uu because differentiating it lowers its degree.

  5. Differentiate uu and integrate dvdx\frac{dv}{dx}

    dudx=1,v=−e−2x2\frac{du}{dx}=1,\quad v=-\frac{e^{-2x}}{2}

    Each application of the formula reduces the degree of the polynomial by one.

  6. Apply the by-parts formula

    ∫xe−2x dx=−xe−2x2−∫−e−2x2 dx\int xe^{-2x}\,dx=-\frac{xe^{-2x}}{2}-\int -\frac{e^{-2x}}{2}\,dx

    The integral that is left has a polynomial of lower degree, so progress has been made.

  7. Complete the integration

    ∫xe−2x dx=(−2x−1)e−2x4+c\int xe^{-2x}\,dx=\frac{\left(-2x-1\right)e^{-2x}}{4}+c

    Collecting every piece gives the antiderivative.

  8. Check the antiderivative by differentiating it

    ddx((−2x−1)e−2x4)=xe−2x\frac{d}{dx}\left(\frac{\left(-2x-1\right)e^{-2x}}{4}\right)=xe^{-2x}

    Differentiating FF must return the integrand exactly; if it does not, the integration is wrong.

  9. Apply the limits of integration

    ∫0∞xe−2x dx=[(−2x−1)e−2x4]0∞\int_{0}^{\infty}xe^{-2x}\,dx=\left[\frac{\left(-2x-1\right)e^{-2x}}{4}\right]_{0}^{\infty}

    The value is the antiderivative at the top minus the antiderivative at the bottom, with a limit taken at any improper endpoint.

  10. Take the limit of the antiderivative as t→∞t\to\infty

    lim⁡t→∞((−2t−1)e−2t4)=0\lim_{t\to\infty}\left(\frac{\left(-2t-1\right)e^{-2t}}{4}\right)=0

    This limit IS the improper integral; no substitution of ∞\infty is ever made.

  11. Evaluate the antiderivative at the lower limit

    F(0)=−14F\left(0\right)=-\frac{1}{4}

    The lower limit is an ordinary number, so it is simply substituted.

  12. Subtract to obtain the value of the integral

    ∫0∞xe−2x dx=(0)−(−14)=14\int_{0}^{\infty}xe^{-2x}\,dx=\left(0\right)-\left(-\frac{1}{4}\right)=\frac{1}{4}

    This is the exact value of the integral.

  13. Check the answer numerically

    I≈0.25I\approx0.25

    High-precision quadrature agrees with the exact value to six significant figures.

  14. Check that the answer is plausible

    xe−2x>0 on the interval⇒I>0xe^{-2x}>0\ \text{on the interval}\Rightarrow I>0

    A positive integrand over an interval of positive width must give a positive answer.

  15. Recall the integral of the reciprocal function

    ∫1x dx=ln⁡∣x∣+c\int\frac{1}{x}\,dx=\ln\left|x\right|+c

    The reciprocal is the single power the power rule cannot handle.

  16. Select the statement that matches the value found

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

    The defining limit is finite, so the improper integral converges to this value.

Answer
The integral converges to 14\text{The integral converges to }\frac{1}{4}
Question 5
9 markschallenging
Find ∫x2+3(x+1)(x2+1) dx\int \frac{x^{2}+3}{\left(x+1\right)\left(x^{2}+1\right)}\,dx.
Show worked solution

Worked solution

  1. Write down the integral to be found

    I=∫x2+3(x+1)(x2+1) dxI=\int \frac{x^{2}+3}{\left(x+1\right)\left(x^{2}+1\right)}\,dx

    There are no limits, so the answer is a family of functions.

  2. Factorise the denominator

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

    The denominator must be a product of irreducible factors before it can be split up.

  3. Write the integrand in partial fractions

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

    Each factor contributes one term for every power up to its multiplicity; a quadratic factor carries a linear numerator.

  4. Multiply through by the denominator

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

    Clearing the fractions leaves an identity that holds for every xx.

  5. Substitute x=−1x=-1 to find AA

    4=A×(2)⇒A=24=A\times\left(2\right)\quad\Rightarrow\quad A=2

    Choosing the root of a factor kills every other term of the identity at once.

  6. Compare coefficients to find the remaining constants

    B=−1,C=1B=-1,\quad C=1

    Equating the coefficients of each power of xx on both sides of the identity gives enough equations for the constants left over.

  7. State the partial fraction form

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

    This is the integrand written as a sum of standard fractions.

  8. Split the term over the quadratic factor

    px+qx2+a2=p2⋅2xx2+a2+qx2+a2\frac{px+q}{x^{2}+a^{2}}=\frac{p}{2}\cdot\frac{2x}{x^{2}+a^{2}}+\frac{q}{x^{2}+a^{2}}

    The first piece is a logarithm and the second is an inverse tangent.

  9. Check the antiderivative by differentiating it

    ddx(2ln⁡(x+1)−ln⁡(x2+1)2+arctan⁡(x))=x2+3(x+1)(x2+1)\frac{d}{dx}\left(2\ln{\left(x+1\right)}-\frac{\ln{\left(x^{2}+1\right)}}{2}+\arctan{\left(x\right)}\right)=\frac{x^{2}+3}{\left(x+1\right)\left(x^{2}+1\right)}

    Differentiating the answer must return the integrand exactly.

  10. Do not forget the constant of integration

    +c+c

    An indefinite integral is only determined up to an additive constant.

  11. Recall the power rule for integration

    ∫xn dx=xn+1n+1+c,n≠−1\int x^{n}\,dx=\frac{x^{n+1}}{n+1}+c,\quad n\neq-1

    Every index except −1-1 is integrated by raising the index by one.

  12. Recall the integral of the reciprocal function

    ∫1x dx=ln⁡∣x∣+c\int\frac{1}{x}\,dx=\ln\left|x\right|+c

    The reciprocal is the single power the power rule cannot handle.

  13. Recall the integral of an exponential

    ∫ekx dx=1kekx+c\int e^{kx}\,dx=\frac{1}{k}e^{kx}+c

    Dividing by the coefficient of xx in the index undoes the chain rule.

  14. Recall the integration by parts formula

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

    Integration by parts is the product rule read backwards.

  15. State the antiderivative

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

    This is the general antiderivative of the integrand.

Answer
2ln⁡(x+1)−ln⁡(x2+1)2+arctan⁡(x)+c2\ln{\left(x+1\right)}-\frac{\ln{\left(x^{2}+1\right)}}{2}+\arctan{\left(x\right)}+c

Unlock 29 more Further calculus questions

Create a free account to work through every Further Maths Further calculus 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 Further calculus practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Argand diagramsComplex numbers: arithmeticModelling with differential equationsComplex numbers: de MoivreFirst-order differential equationsHyperbolic functions