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### Answers (1)

Answer: $-\frac{\log x}{(x+1)}+\log x-\log (x+1)$

Hint:

Given: $\int \frac{log\: x}{\left ( x+1 \right )^{2}}dx$

Solution:

$I= \frac{log\: x}{\left ( x+1 \right )^{2}}dx$

$=log\: x\int \frac{1}{\left ( x+1 \right )^{2}}dx-\int \left ( \frac{1}{x}\int \frac{1}{\left ( x+1 \right )^{2}} dx\right )dx$

$=log\: x\left ( \frac{-1}{x+1} \right )+\int \frac{1}{x}\frac{1}{x+1}dx$

$=-\frac{log\: x}{\left ( x+1 \right )}+I_{1}$

$=I_{1}=\int \frac{1}{x\left ( x+1 \right )}dx$

Using partial fractions

Let $\frac{1}{x\left ( x+1 \right )}=\frac{A}{x}+\frac{B}{x+1}$

$1=A\left ( x+1 \right )+B\left ( x \right )$

$1=Ax+A+Bx$

$1=\left ( A+B \right )x+A$

$\Rightarrow A+B=0$

$\Rightarrow A=1$

$\Rightarrow B=-1$

$I_{1}=\int \left [ \frac{1}{x} -\frac{1}{x+1}\right ]dx$

$=log\: x-log\left ( x+1 \right )$

$I=\frac{log\, x}{x+1}+log\: x-log\left ( x+1 \right )$

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