Pre-U Pre-U 9794/1 2012 June — Question 11 11 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2012
SessionJune
Marks11
TopicIntegration by Parts
TypeSequential multi-part (building on previous)
DifficultyStandard +0.3 Part (i) is a standard integration by parts demonstration that appears in most textbooks. Part (ii)(a) requires applying integration by parts twice, which is routine once the pattern is recognized. Part (ii)(b) is a straightforward substitution (u = ln x) after recognizing the derivative pattern. While multi-part, each component uses standard techniques without requiring novel insight, making this slightly easier than average.
Spec1.08i Integration by parts
11
  1. Use integration by parts to show that \(\int \ln x \mathrm {~d} x = x \ln x - x + c\).
  2. Find
    1. \(\int ( \ln x ) ^ { 2 } \mathrm {~d} x\),
    2. \(\quad \int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x\).
Part (i)
- Use \(f' = 1\) and \(g = \ln x\) and apply the correct formula for integration by parts: M1
- Obtain correctly \(\int \ln x\, dx = x\ln x - x + c\) AG: A1 [2]
Part (ii)(a)
*Method 1* (Integration by parts using \((\ln x)^2\) as \(f' = \ln x\) and \(g = \ln x\)):
- Obtain \((\ln x)(x\ln x - x) - \int f(x)dx\): B1
- Obtain \(g(x) - \int\dfrac{x\ln x - x}{x}dx\): B1
- Attempt to simplify integral and substitute result from (i): M1
- Obtain \(\int(\ln x - 1)dx = x\ln x - x - x\) and hence \(x(\ln x)^2 - 2\ln x + 2x\) (+c): A1
*Method 2* (Integration by parts using \((\ln x)^2\) as \(1 \times (\ln x)^2\)):
- Obtain \(x(\ln x)^2 - \int f(x)dx\): B1
- Obtain \(g(x) - \int\dfrac{2x\ln x}{x}dx\): B1
- Attempt to simplify integral and substitute result from (i): M1
- Obtain \(2\int\ln x\,dx = 2(x\ln x - x)\) and hence \(x(\ln x)^2 - 2x\ln x + 2x\) (+c): A1
*Method 3* (Integration by parts twice using \((\ln x)^2 = u^2\)):
- Obtain \(u^2 e^u - \int f(x)dx\): B1
- Obtain \(g(x) - \int 2ue^u\,du\): B1
- Attempt to integrate again: M1
- Obtain \(\int 2ue^u\,du = 2(ue^u - e^u)\) and hence \(x(\ln x)^2 - 2x\ln x + 2x\) (+c): A1 [4]
Part (ii)(b)
*Method 1* (Using parts):
- Attempt integration by parts as \(g(x) - \int f(x)dx\): M1
- Obtain \((\ln x)(\ln(\ln x)) - \int f(x)dx\): 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
*Method 2* (Using substitution):
- Attempt to obtain an integral in \(u\) by stating or implying \(u = \ln x\) AND \(du = \dfrac{1}{x}dx\) OR \(u = \ln x\) AND \(x = e^u\) AND \(dx = e^u\,du\): M1
- Obtain directly \(\int\ln u\,du\) OR \(\int\dfrac{\ln u}{e^u}e^u\,du\) and cancel to obtain \(\int\ln u\,du\): A1
- Obtain \(u(\ln u) - u\): A1
- Obtain \((\ln x)(\ln(\ln x)) - \ln x\) (+c): A1
- Use \(+c\) in (b)(i) and (ii): B1 [5]
[Total: 11]
**Part (i)**
- Use $f' = 1$ and $g = \ln x$ and apply the correct formula for integration by parts: M1
- Obtain correctly $\int \ln x\, dx = x\ln x - x + c$ AG: A1 **[2]**

**Part (ii)(a)**

*Method 1* (Integration by parts using $(\ln x)^2$ as $f' = \ln x$ and $g = \ln x$):
- Obtain $(\ln x)(x\ln x - x) - \int f(x)dx$: B1
- Obtain $g(x) - \int\dfrac{x\ln x - x}{x}dx$: B1
- Attempt to simplify integral and substitute result from (i): M1
- Obtain $\int(\ln x - 1)dx = x\ln x - x - x$ and hence $x(\ln x)^2 - 2\ln x + 2x$ (+c): A1

*Method 2* (Integration by parts using $(\ln x)^2$ as $1 \times (\ln x)^2$):
- Obtain $x(\ln x)^2 - \int f(x)dx$: B1
- Obtain $g(x) - \int\dfrac{2x\ln x}{x}dx$: B1
- Attempt to simplify integral and substitute result from (i): M1
- Obtain $2\int\ln x\,dx = 2(x\ln x - x)$ and hence $x(\ln x)^2 - 2x\ln x + 2x$ (+c): A1

*Method 3* (Integration by parts twice using $(\ln x)^2 = u^2$):
- Obtain $u^2 e^u - \int f(x)dx$: B1
- Obtain $g(x) - \int 2ue^u\,du$: B1
- Attempt to integrate again: M1
- Obtain $\int 2ue^u\,du = 2(ue^u - e^u)$ and hence $x(\ln x)^2 - 2x\ln x + 2x$ (+c): A1 **[4]**

**Part (ii)(b)**

*Method 1* (Using parts):
- Attempt integration by parts as $g(x) - \int f(x)dx$: M1
- Obtain $(\ln x)(\ln(\ln x)) - \int f(x)dx$: 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

*Method 2* (Using substitution):
- Attempt to obtain an integral in $u$ by stating or implying $u = \ln x$ AND $du = \dfrac{1}{x}dx$ OR $u = \ln x$ AND $x = e^u$ AND $dx = e^u\,du$: M1
- Obtain directly $\int\ln u\,du$ OR $\int\dfrac{\ln u}{e^u}e^u\,du$ and cancel to obtain $\int\ln u\,du$: A1
- Obtain $u(\ln u) - u$: A1
- Obtain $(\ln x)(\ln(\ln x)) - \ln x$ (+c): A1
- Use $+c$ in **(b)(i)** and **(ii)**: B1 **[5]**

**[Total: 11]**
11 (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 $\int ( \ln x ) ^ { 2 } \mathrm {~d} x$,
\item $\quad \int \frac { \ln ( \ln x ) } { x } \mathrm {~d} x$.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2012 Q11 [11]}}
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