2022TrialShort answer

Q16 (c)

3 marks

Question

Find
$\int\frac{\ln x-1}{(x+\ln x)^2}\,dx.$

Try the question before revealing the answer.

Solution

I=\int\frac{\ln x-1}{(x+\ln x)^2}\,dx

I=\int\frac{x+\ln x-x-1}{(x+\ln x)^2}\,dx

Let

J=\int\frac{x\left(1+\frac1x\right)}{(x+\ln x)^2}\,dx.

By using integration by parts on $J$ ,

\begin{aligned} I&=\int\frac{x+\ln x}{(x+\ln x)^2}\,dx-\int\frac{x+1}{(x+\ln x)^2}\,dx\\ &=\int\frac1{x+\ln x}\,dx-J. \end{aligned}

\begin{aligned} J&=u\,v-\int v\,du\\ u&=x,\qquad v' =\frac{1+\frac1x}{(x+\ln x)^2},\\ u'&=1,\qquad v=-\frac1{x+\ln x}\quad\text{by Reverse Chain Rule}\\ J&=x\left(-\frac1{x+\ln x}\right)-\int\frac{-1}{x+\ln x}\,dx\\ &=-\frac{x}{x+\ln x}+\int\frac1{x+\ln x}\,dx. \end{aligned}

Therefore,

\begin{aligned} I&=\int\frac1{x+\ln x}\,dx-\left(-\frac{x}{x+\ln x}+\int\frac1{x+\ln x}\,dx\right)\\ &=\frac{x}{x+\ln x}+C. \end{aligned}

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Calculus

Further IntegrationApply substitution, partial fractions, integration by parts and recurrence relationships to theoretical and practical problems.

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