| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/1 (Pre-U Mathematics Paper 1) |
| Year | 2019 |
| Session | Specimen |
| Marks | 2 |
| Topic | Integration by Parts |
| Type | Sequential multi-part (building on previous) |
| Difficulty | Standard +0.8 Part (a) is a standard integration by parts exercise. Part (b)(i) requires applying the technique twice with careful algebraic manipulation. Part (b)(ii) requires recognizing a substitution (u = ln x) combined with the result from (a), 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 |
(a) Use $f' = 1$ and $g = \ln x$ and apply the correct formula for integration by parts — **M1**
Obtain **AG** correctly — **A1**
**Total: 2**
(b)(i) $f' = \ln x$ and $g = \ln x$ — **B1**
Obtain $(\ln x)(x\ln x - x) - \int \text{f}(x)\,dx$ — **B1**
Attempt to simplify integral and substitute result from **(a)** — **M1**
Obtain $\int(\ln x - 1)\,dx = x\ln x - x - x$ and hence $x(\ln x)^2 - 2x\ln x + 2x\ (+c)$ — **A1**
**Total: 4**
(b)(ii) Attempt integration by parts as $g(x) - \int \text{f}(x)\,dx$ — **M1**
Obtain $(\ln x)(\ln(\ln x)) - \int \text{f}(x)\,dx$ — **A1**
Obtain $g(x) - \int \dfrac{1}{x}\,dx$ — **A1**
Obtain $(\ln x)(\ln(\ln x)) - \ln x + c$ — **A1**
Sight of $+c$ in last two parts — **B1**
**Total: 5**12
\begin{enumerate}[label=(\alph*)]
\item Use integration by parts to show that $\int \ln x \mathrm {~d} x = x \ln x - x + c$.
\item Find
\begin{enumerate}[label=(\roman*)]
\item $\quad \int ( \ln x ) ^ { 2 } \mathrm {~d} x$,
\item $\quad \int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2019 Q12 [2]}}