AQA C3 — Question 10

Exam BoardAQA
ModuleC3 (Core Mathematics 3)
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeSequential multi-part (building on previous)
DifficultyStandard +0.3 This is a standard C3 integration question with routine techniques. Part (a)(i) is a textbook example of integration by parts with ln x; part (a)(ii) applies the same method twice; part (b) is a straightforward substitution with clear guidance. All steps are mechanical applications of standard methods with no novel insight required, making it slightly easier than average.
Spec1.08h Integration by substitution1.08i Integration by parts
10
    1. By writing \(\ln x\) as \(( \ln x ) \times 1\), use integration by parts to find \(\int \ln x \mathrm {~d} x\).
    2. Find \(\int ( \ln x ) ^ { 2 } \mathrm {~d} x\).
  2. Use the substitution \(u = \sqrt { x }\) to find the exact value of $$\int _ { 1 } ^ { 4 } \frac { 1 } { x + \sqrt { x } } \mathrm {~d} x$$ (7 marks)
Question 10:
(a)(i) \(\int \ln x\,dx\): let \(u=\ln x\), \(dv=dx\); \(du=\frac{1}{x}dx\), \(v=x\)
AnswerMarks
\(= x\ln x - \int 1\,dx = x\ln x - x + c\)M1 parts, A1 \(x\ln x\), A1 \(-x\), A1 \(+c\)
(a)(ii) \(\int(\ln x)^2\,dx\): let \(u=(\ln x)^2\), \(dv=dx\)
AnswerMarks
\(= x(\ln x)^2 - 2\int \ln x\,dx = x(\ln x)^2 - 2x\ln x + 2x + c\)M1 parts, A1 \(x(\ln x)^2\), M1 using (i), A1 final answer
(b) \(u=\sqrt{x}\), \(u^2=x\), \(2u\,du=dx\); limits \(x=1\to u=1\), \(x=4\to u=2\)
\[\int_1^4\frac{1}{x+\sqrt{x}}dx = \int_1^2\frac{2u}{u^2+u}du = \int_1^2\frac{2}{u+1}du = \left[2\ln(u+1)\right]_1^2 = 2\ln 3 - 2\ln 2 = 2\ln\frac{3}{2}\]
AnswerMarks
M1 substitution, A1 \(dx=2u\,du\), M1 simplifying integrand, A1 \(\frac{2}{u+1}\), M1 integrating, A1 \(2\ln(u+1)\), A1 \(2\ln\frac{3}{2}\) 
## Question 10:

**(a)(i)** $\int \ln x\,dx$: let $u=\ln x$, $dv=dx$; $du=\frac{1}{x}dx$, $v=x$

$= x\ln x - \int 1\,dx = x\ln x - x + c$ | **M1** parts, **A1** $x\ln x$, **A1** $-x$, **A1** $+c$

**(a)(ii)** $\int(\ln x)^2\,dx$: let $u=(\ln x)^2$, $dv=dx$

$= x(\ln x)^2 - 2\int \ln x\,dx = x(\ln x)^2 - 2x\ln x + 2x + c$ | **M1** parts, **A1** $x(\ln x)^2$, **M1** using (i), **A1** final answer

**(b)** $u=\sqrt{x}$, $u^2=x$, $2u\,du=dx$; limits $x=1\to u=1$, $x=4\to u=2$

$$\int_1^4\frac{1}{x+\sqrt{x}}dx = \int_1^2\frac{2u}{u^2+u}du = \int_1^2\frac{2}{u+1}du = \left[2\ln(u+1)\right]_1^2 = 2\ln 3 - 2\ln 2 = 2\ln\frac{3}{2}$$

| **M1** substitution, **A1** $dx=2u\,du$, **M1** simplifying integrand, **A1** $\frac{2}{u+1}$, **M1** integrating, **A1** $2\ln(u+1)$, **A1** $2\ln\frac{3}{2}$
10
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item By writing $\ln x$ as $( \ln x ) \times 1$, use integration by parts to find $\int \ln x \mathrm {~d} x$.
\item Find $\int ( \ln x ) ^ { 2 } \mathrm {~d} x$.
\end{enumerate}\item Use the substitution $u = \sqrt { x }$ to find the exact value of

$$\int _ { 1 } ^ { 4 } \frac { 1 } { x + \sqrt { x } } \mathrm {~d} x$$

(7 marks)
\end{enumerate}

\hfill \mbox{\textit{AQA C3  Q10}}
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