Integration — Worked Solutions (HSC Maths Extension 2)
Created with Intu AI Reviewed by Intuition's expert tutors
Studying this? See our HSC Maths Extension 2 course →
Worked examples for HSC Maths Extension 2 integration. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Extension 2 — Integration. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
The first decision in any integral is which technique — recognise a partial-fraction candidate by its factorable rational denominator, and a by-parts candidate by a product of two different function types.
Example 1 — Partial fractions
Question
Find $\displaystyle\int \dfrac{5x - 1}{(x - 1)(x + 2)}\,dx$.
Solution
A proper rational function with distinct linear factors — split it into partial fractions.
Write $\dfrac{5x - 1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$, so $5x - 1 = A(x+2) + B(x-1)$.
Let $x = 1$: $4 = 3A \Rightarrow A = \tfrac43$. Let $x = -2$: $-11 = -3B \Rightarrow B = \tfrac{11}{3}$.
Then $\displaystyle\int \left(\dfrac{4/3}{x-1} + \dfrac{11/3}{x+2}\right)dx = \dfrac43\ln|x-1| + \dfrac{11}{3}\ln|x+2| + C$.
Use the substitution-of-roots shortcut to find $A$ and $B$ fast — don't expand and equate coefficients unless you have to.
The denominator is already factored into two distinct linear pieces, which is the signal for partial fractions — we rewrite one awkward fraction as a sum of two simple ones we can integrate.
Set $\dfrac{5x - 1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$. Multiplying through by the denominator gives $5x - 1 = A(x+2) + B(x-1)$.
The neat trick is to substitute the values that kill one bracket. Putting $x = 1$ makes the $B$ term vanish: $5(1) - 1 = A(3)$, so $4 = 3A$ and $A = \tfrac43$. Putting $x = -2$ kills the $A$ term: $5(-2) - 1 = B(-3)$, so $-11 = -3B$ and $B = \tfrac{11}{3}$.
Now each piece integrates to a logarithm because each is a constant over a linear term: $\displaystyle\int\dfrac{4/3}{x-1}\,dx + \int\dfrac{11/3}{x+2}\,dx = \dfrac43\ln|x-1| + \dfrac{11}{3}\ln|x+2| + C$. The absolute-value signs matter — the log is only defined for positive arguments.
Decompose into partial fractions, then integrate each term.
- $\dfrac{5x-1}{(x-1)(x+2)} = \dfrac{A}{x-1} + \dfrac{B}{x+2}$
- $5x - 1 = A(x+2) + B(x-1)$
- $x = 1:\ 4 = 3A \Rightarrow A = \tfrac43$
- $x = -2:\ -11 = -3B \Rightarrow B = \tfrac{11}{3}$
- $\displaystyle\int = \dfrac43\ln|x-1| + \dfrac{11}{3}\ln|x+2| + C$
Where the marks go
- 1 mark: Correct partial-fraction setup $5x - 1 = A(x+2) + B(x-1)$
- 1 mark: Correct constants $A = \frac43$ and $B = \frac{11}{3}$
- 1 mark: Correct integral $\frac43\ln|x-1| + \frac{11}{3}\ln|x+2| + C$
Key idea
A proper rational function with distinct linear factors splits as $\sum \dfrac{\text{const}}{\text{linear}}$; substituting each root isolates one constant at a time, and each term integrates to a logarithm.
Example 2 — Integration by parts
Question
Find $\displaystyle\int x^2 \ln x \,dx$.
Solution
Product of a power and a log — integrate by parts, $\displaystyle\int u\,dv = uv - \int v\,du$.
Choose $u = \ln x$ (it simplifies when differentiated) and $dv = x^2\,dx$. Then $du = \dfrac{1}{x}\,dx$ and $v = \dfrac{x^3}{3}$.
$\displaystyle\int x^2\ln x\,dx = \dfrac{x^3}{3}\ln x - \int \dfrac{x^3}{3}\cdot\dfrac{1}{x}\,dx = \dfrac{x^3}{3}\ln x - \dfrac13\int x^2\,dx$.
The remaining integral is easy: $\dfrac13\cdot\dfrac{x^3}{3} = \dfrac{x^3}{9}$. So the answer is $\dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$.
Pick $u = \ln x$ — differentiating the log is what makes the second integral tractable. The reverse choice goes nowhere.
We've got a product of two different kinds of function — a power $x^2$ and a logarithm $\ln x$ — which is the classic signal for integration by parts: $\displaystyle\int u\,dv = uv - \int v\,du$.
The art is choosing $u$ and $dv$ well. We want $u$ to get simpler when differentiated, and $\ln x$ does exactly that (its derivative is $\tfrac1x$, no more log). So let $u = \ln x$ and $dv = x^2\,dx$. Then $du = \dfrac{1}{x}\,dx$ and, integrating, $v = \dfrac{x^3}{3}$.
Substituting in: $\displaystyle\int x^2\ln x\,dx = \dfrac{x^3}{3}\ln x - \int \dfrac{x^3}{3}\cdot\dfrac{1}{x}\,dx$. Notice how the log has disappeared from the leftover integral — that's the whole point.
Simplify the remaining integrand: $\dfrac{x^3}{3}\cdot\dfrac1x = \dfrac{x^2}{3}$, and $\displaystyle\int \dfrac{x^2}{3}\,dx = \dfrac{x^3}{9}$. Putting it together, $\dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$.
By parts, $\displaystyle\int u\,dv = uv - \int v\,du$.
- $u = \ln x,\ du = \tfrac1x\,dx$
- $dv = x^2\,dx,\ v = \tfrac{x^3}{3}$
- $\displaystyle\int x^2\ln x\,dx = \dfrac{x^3}{3}\ln x - \int \dfrac{x^2}{3}\,dx$
- $= \dfrac{x^3}{3}\ln x - \dfrac{x^3}{9} + C$
Where the marks go
- 1 mark: Chooses $u = \ln x$, $dv = x^2\,dx$ (suitable parts)
- 1 mark: Correct $du = \frac1x\,dx$ and $v = \frac{x^3}{3}$
- 1 mark: Applies the by-parts formula and simplifies the remaining integrand to $\frac{x^2}{3}$
- 1 mark: Correct final answer $\frac{x^3}{3}\ln x - \frac{x^3}{9} + C$
Key idea
For a product of unlike functions use integration by parts; choose $u$ as the factor that simplifies when differentiated (here $\ln x$) so the remaining integral is easier.
Frequently asked questions
Step-by-step solutions to exam-style questions on Integration in HSC Maths Extension 2, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Extension 2 syllabus for Integration, using the methods and notation expected in the exam.