| Exam Board | WJEC |
|---|---|
| Module | Further Unit 4 (Further Unit 4) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Partial fractions with irreducible quadratic |
| Difficulty | Standard +0.8 This is a Further Maths question requiring partial fractions with an irreducible quadratic factor (requiring complex roots or cover-up method extensions), followed by integration involving both logarithmic and arctangent forms. While the techniques are standard for Further Maths, the algebraic manipulation is non-trivial and the integration of the arctangent term requires careful handling. The 10-mark allocation and multi-step nature place it moderately above average difficulty. |
| Spec | 1.08j Integration using partial fractions4.05c Partial fractions: extended to quadratic denominators |
| Answer | Marks | Guidance |
|---|---|---|
| Let \(\frac{8x^2 + x + 5}{(2x+1)(x^2+3)} = \frac{A}{2x+1} + \frac{Bx+C}{x^2+3}\) | M1 | AO1 |
| \(= \frac{A(x^2+3) + (Bx+C)(2x+1)}{(2x+1)(x^2+3)}\) | A1 | AO1 |
| \(A = 2, B = 3, C = -1\) | A1, A1 | AO1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{Integral} = \int_2^3 \left(\frac{2}{2x+1} + \frac{3x}{x^2+3} - \frac{1}{x^2+3}\right)dx\) | M1 | AO1 |
| \(= \left[\ln(2x+1) + \frac{3}{2}\ln(x^2+3) - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)\right]_2^3\) | A1, A1 | AO1 |
| \(= \ln 7 + \frac{3}{2}\ln 12 - \frac{1}{\sqrt{3}}\tan^{-1}\sqrt{3} - \ln 5 - \frac{3}{2}\ln 7 + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)\) | A1 | AO1 |
| \(= 1.035\) | A1 | AO1 |
## Part (a)
Let $\frac{8x^2 + x + 5}{(2x+1)(x^2+3)} = \frac{A}{2x+1} + \frac{Bx+C}{x^2+3}$ | M1 | AO1
$= \frac{A(x^2+3) + (Bx+C)(2x+1)}{(2x+1)(x^2+3)}$ | A1 | AO1
$A = 2, B = 3, C = -1$ | A1, A1 | AO1
## Part (b)
$\text{Integral} = \int_2^3 \left(\frac{2}{2x+1} + \frac{3x}{x^2+3} - \frac{1}{x^2+3}\right)dx$ | M1 | AO1
$= \left[\ln(2x+1) + \frac{3}{2}\ln(x^2+3) - \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x}{\sqrt{3}}\right)\right]_2^3$ | A1, A1 | AO1
$= \ln 7 + \frac{3}{2}\ln 12 - \frac{1}{\sqrt{3}}\tan^{-1}\sqrt{3} - \ln 5 - \frac{3}{2}\ln 7 + \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$ | A1 | AO1
$= 1.035$ | A1 | AO1
**Total: [10]**
---The function $f$ is defined by
$$f(x) = \frac{8x^2 + x + 5}{(2x + 1)(x^2 + 3)}.$$
\begin{enumerate}[label=(\alph*)]
\item Express $f(x)$ in partial fractions. [4]
\item Hence evaluate
$$\int_2^5 f(x)dx,$$
giving your answer correct to three decimal places. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 4 Q7 [10]}}