Worked Solutions
Integration — Worked Solutions (HSC Maths Extension 2)
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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.
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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.