Specialist Mathematics · Mock exam 2026 All mock exam questions
Mock exam · Paper 1 · Simple familiar

Question 14

[4 marks] Technology-free
The question

In this question, $x>0$.

a)

Express $\dfrac{1}{x(x+1)}$ in partial fractions.

[1 mark] b)

Use integration by parts and your result from Question 14a) to determine $\displaystyle\int\frac{\ln(x)}{(x+1)^{2}}\,dx.$

[3 marks]
Watch the situation first

Part a) builds the piece that part b) needs

Part a) looks like a warm-up. Watch the fraction it gives you turn up again in the middle of the integration by parts.

Part a): split the fraction into two simpler fractionsx=0 clears the B term, and x=−1 clears the A termSo the fraction is a difference of two simple fractionsPart b): differentiate ln(x), integrate the other factorv times the derivative of u is the fraction from part a)Two minus signs multiply to give a plusSwap in part a), then integrate each term1x(x+1)=Ax+Bx+11=A(x+1)+Bxx=0:A=1x=−1:B=−11x(x+1)=1x−1x+1From part a)∫ln(x)(x+1)2dxu=ln(x)dvdx=1(x+1)2differentiateintegratedudx=1xv=−1x+1v·dudx=−1x+1·1x=−1x(x+1)∫ln(x)(x+1)2dx=−ln(x)x+1−∫(−1x+1)·1xdx=−ln(x)x+1+∫1x(x+1)dx∫ln(x)(x+1)2dx=−ln(x)x+1+ln(x)−ln(x+1)+c
Now for the mathematics

Find A and B, then follow the signs

Substitute the two values of $x$ that each clear a term. Then choose $v$ and tidy the sign in front of the new integral. The readout shows what each choice gives you.

uv∫vdudxdxdifferentiateintegrateu=ln(x)dvdx=1(x+1)2dudx=1xv=?v=−1x+1v=1x+1From part a)A= ?,B= ?1x(x+1)=Ax+Bx+11x(x+1)=1x−1x+1
Part a) · substitute into $1=A(x+1)+Bx$
Part b) · choose $v$, an antiderivative of $\dfrac{1}{(x+1)^{2}}$
Tidy the sign in front of the new integral
Part a)
Part b)

Start with part a). Pick a value of $x$ that makes one term vanish.

One constant found. Substitute the other value to find the second.

Part a) is done. Now choose $v$.

Integrating $(x+1)^{-2}$ gives $\dfrac{(x+1)^{-1}}{-1}=-\dfrac{1}{x+1}$, so the minus sign belongs in $v$. Carried through, your answer is the exact negative of the right one. Mark 2 is lost.

Good. Now tidy the sign in front of the new integral.

You dropped the double negative. The minus in front of the integral and the minus inside $v$ multiply to give a plus. Mark 3 is lost.

Finish part a) to integrate the last term.

That’s the integral. Part a) turns the last integral into $\ln(x)-\ln(x+1)+c$, so the answer is $-\dfrac{\ln(x)}{x+1}+\ln(x)-\ln(x+1)+c$. There is no need to simplify further.

Marking guide · 4 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it.

Choose a method

Step 1 · part a)1 mark

Write one constant over each linear factor, clear the denominators, then substitute $x=0$ and $x=-1$.

$$\begin{aligned}\frac{1}{x(x+1)}&=\frac{A}{x}+\frac{B}{x+1}\\1&=A(x+1)+Bx\\x=0\text{:}\quad A&=1\\x=-1\text{:}\quad B&=-1\end{aligned}$$

Why?

$1=A(x+1)+Bx$ is an equation between polynomials that holds for every $x>0$, so it holds for every $x$. You may substitute $x=0$ and $x=-1$ even though they are outside the domain.

$\dfrac{1}{x(x+1)}=\dfrac{1}{x}-\dfrac{1}{x+1}$

Marker: correctly determines the partial fractions

Step 2 · part b)1 mark

Let $u=\ln(x)$ and $\dfrac{dv}{dx}=\dfrac{1}{(x+1)^2}$, then differentiate $u$ and integrate $\dfrac{dv}{dx}$.

$$\begin{aligned}u=\ln(x)\ &\Rightarrow\ \frac{du}{dx}=\frac{1}{x}\\\frac{dv}{dx}=\frac{1}{(x+1)^2}\ &\Rightarrow\ v=-\frac{1}{x+1}\end{aligned}$$

Why?

Differentiating $\ln(x)$ gives the simple term $\dfrac{1}{x}$, and $(x+1)^{-2}$ integrates directly by the power rule. Then $v\,\dfrac{du}{dx}$ is a multiple of $\dfrac{1}{x(x+1)}$, the 14a) expression.

Marker: correctly determines $\dfrac{du}{dx}$ and $v$

Step 3 · part b)1 mark

Substitute into the integration by parts formula and tidy the signs.

$$\begin{aligned}&\int\frac{\ln(x)}{(x+1)^2}\,dx\\&\quad=-\frac{\ln(x)}{x+1}-\int\left(-\frac{1}{x+1}\right)\frac{1}{x}\,dx\\&\quad=-\frac{\ln(x)}{x+1}+\int\frac{1}{x(x+1)}\,dx\end{aligned}$$

Why?

The minus in front of the integral and the minus inside $v$ multiply to give a plus. This is the most common place to lose a sign.

Marker: correctly applies the integration by parts formula

Step 4 · part b)1 mark

Replace $\dfrac{1}{x(x+1)}$ with $\dfrac{1}{x}-\dfrac{1}{x+1}$ from 14a) and integrate each term. Since $x>0$, no absolute value bars are needed.

$$\begin{aligned}&\int\frac{\ln(x)}{(x+1)^2}\,dx\\&\quad=-\frac{\ln(x)}{x+1}+\ln(x)-\ln(x+1)+c\end{aligned}$$

$\displaystyle\int\frac{\ln(x)}{(x+1)^2}\,dx=$ $\displaystyle-\frac{\ln(x)}{x+1}+\ln(x)-\ln(x+1)+c$
Marking

Equivalent forms are accepted, for example $\ln\left(\dfrac{x}{x+1}\right)-\dfrac{\ln(x)}{x+1}+c$. Follow-through applies in Question 14b) from an incorrect result in Question 14a).

Marker: determines the integral using the result from Question 14a)

Putting it all together

Each phrase of the question gave you something

“In this question, $x>0$”
Both $x$ and $x+1$ are positive, so $\ln|x|$ and $\ln|x+1|$ lose their absolute value bars.
“Express $\dfrac{1}{x(x+1)}$ in partial fractions”
Two different linear factors, so one constant over each: $\dfrac{A}{x}+\dfrac{B}{x+1}$. Then $x=0$ and $x=-1$ give $A=1$ and $B=-1$.
“Use integration by parts”
The integrand is a product. Differentiate $\ln(x)$ to get $\dfrac{1}{x}$, and integrate $(x+1)^{-2}$ to get $v=-\dfrac{1}{x+1}$.
“and your result from Question 14a)”
It confirms the choice of $u$: $v\,\dfrac{du}{dx}$ is a multiple of $\dfrac{1}{x(x+1)}$, and part a) turns that into two standard logarithm integrals.
“determine $\displaystyle\int\frac{\ln(x)}{(x+1)^{2}}\,dx$”
An indefinite integral, with $+c$: $-\dfrac{\ln(x)}{x+1}+\ln(x)-\ln(x+1)+c$. There is no need to simplify further.
Common errors

If I had to guess where you might lose marks…

  • A sign error in $v$. Taking $v=+\dfrac{1}{x+1}$ gives $\dfrac{\ln(x)}{x+1}-\ln(x)+\ln(x+1)+c$, the exact negative of the answer. You lose Mark 2.
  • Dropping the double negative. Writing $-\dfrac{\ln(x)}{x+1}-\displaystyle\int\frac{1}{x(x+1)}\,dx$ gives $-\dfrac{\ln(x)}{x+1}-\ln(x)+\ln(x+1)+c$. You lose Mark 3.
  • The wrong sign on $B$ in a). $\dfrac{1}{x}+\dfrac{1}{x+1}$ recombines to $\dfrac{2x+1}{x(x+1)}$, not $\dfrac{1}{x(x+1)}$. Carried into b), it gives $-\dfrac{\ln(x)}{x+1}+\ln(x)+\ln(x+1)+c$. Part a) earns 0, but follow-through applies in b).
  • Differentiating instead of integrating $\dfrac{dv}{dx}$. Writing $v=-\dfrac{2}{(x+1)^{3}}$ (the derivative of $(x+1)^{-2}$) makes the second integral harder, not easier, and no correct answer follows.
  • Bad log algebra when you simplify. $\ln(x)-\ln(x+1)$ is $\ln\left(\dfrac{x}{x+1}\right)$, not $\dfrac{\ln(x)}{\ln(x+1)}$. If you simplify when you don’t need to, you can lose a mark you had already earned.

Only mark this done when you could do it without help. Reading the solution does not count.

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A Tangent Tuition mock exam question. It is not a past QCAA question, and Tangent Tuition is not affiliated with the QCAA.

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