| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2016 |
| Session | Specimen |
| Marks | 11 |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Standard +0.8 Part (i) is a standard integration by parts exercise. Part (ii)(a) requires applying the technique twice with careful algebraic manipulation. Part (ii)(b) requires recognizing a substitution (u = ln x) combined with the result from (i), showing good problem-solving insight beyond routine application. The sequential building and need for strategic thinking elevates this above average difficulty. |
| Spec | 1.08i Integration by parts |
**Question 12(i)**
- Use $f' = 1$ and $g = \ln x$ and apply the correct formula for integration by parts — M1
- Obtain AG correctly — A1
**Question 12(ii)(a)**
- $f' = \ln x$ and $g = \ln x$ — B1
- Obtain $(\ln x)(x\ln x - x) - \int \text{f}(x)\,\text{d}x$ — B1
- Attempt to simplify integral and substitute result from **(i)** — M1
- Obtain $\int(\ln x - 1)\,\text{d}x = x\ln x - x - x$ and hence $x(\ln x)^2 - 2x\ln x + 2x\ (+c)$ — A1
**Question 12(ii)(b)**
- Attempt integration by parts as $g(x) - \int \text{f}(x)\,\text{d}x$ — M1
- Obtain $(\ln x)(\ln(\ln x)) - \int \text{f}(x)\,\text{d}x$ — A1
- Obtain $g(x) - \int \dfrac{1}{x}\,\text{d}x$ — A1
- Obtain $(\ln x)(\ln(\ln x)) - \ln x + c$ — A1
- Sight of $+c$ in last two parts — B1
**Total: 11 marks**12 (i) Use integration by parts to show that $\int \ln x \mathrm {~d} x = x \ln x - x + c$.\\
(ii) Find
\begin{enumerate}[label=(\alph*)]
\item $\quad \int ( \ln x ) ^ { 2 } \mathrm {~d} x$,
\item $\quad \int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x$.
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2016 Q12 [11]}}