Question

The value of $\int \frac{ln(1+\frac{1}{x})}{x(x+1)}dx$ is?

Answer 1

we are given that $I=\int \frac{ln(1+\frac{1}{x})}{x(x+1)}dx$

let $ln(1+\frac{1}{x})=t$

now differentiating w.r.t. $x$ . we get ,

$\frac{1}{(1+\frac{1}{x})}\frac{d}{dx}(1+\frac{1}{x})=\frac{dt}{dx}$

$\frac{1}{(1+\frac{1}{x})}(0+\frac{-1}{x^2})=\frac{dt}{dx}$

$\therefore \frac{-1}{\frac{(x+1).x^2}{x}}dx=dt$

$\frac{-1}{x(x+1)}dx=dt$

$so,I=\int t(-dt)$

$I=-\int tdt$

$I=\frac{-t^2}{2}+C$

$I=\frac{-1}{2}\left[ln\right(1+\frac{1}{x}){]}^2+C$