#### provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise  18 .9 question 2

Answer:$-\frac{1}{2}\left[\log \left(1+\frac{1}{x}\right)\right]^{2}+c$

Hint: Use substitution method to solve this integral.

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

Solution:

Let $I=\int \frac{\log \left(1+\frac{1}{x}\right)}{x(1+x)} d x$

Put $\log \left(1+\frac{1}{x}\right)=t \Rightarrow \frac{1}{1+\frac{1}{x}}\left(\frac{-1}{x^{2}}\right) d x=d t$

\begin{aligned} &\Rightarrow \frac{1}{\frac{x+1}{x}}\left(\frac{-1}{x^{2}}\right) d x=d t \\ &\Rightarrow \frac{x}{x+1}\left(\frac{-1}{x^{2}}\right) d x=d t \\ &\Rightarrow \frac{-1}{x(x+1)} d x=d t \Rightarrow d x=-(x+1) x \; d t \text { then } \end{aligned}

$I=\int-\frac{t}{x(1+x)}(x+1) x \; d t=-\int t d t \quad\left[\because \frac{1}{x} d x=d t \Rightarrow d x=x\; d t\right]$

\begin{aligned} &=-\frac{t^{1+1}}{1+1}+c \\ &=\frac{-t^{2}}{2}+c \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=\frac{-1}{2}\left\{\log \left(1+\frac{1}{x}\right)\right\}^{2}+c \quad\left[\because t=\log \left(1+\frac{1}{x}\right)\right] \end{aligned}