Question
Evaluate $I=\int \frac{\log \left(1+\frac{1}{x}\right)}{x(1+x)} d x$
✓
Answer
Let $I=\int \frac{\log \left(1+\frac{1}{x}\right)}{x(1+x)} d x$.
Let $\log \left(1+\frac{1}{x}\right)= t$ then,
$d\left(\log \left(1+\frac{1}{x}\right)\right)= dt$
$\Rightarrow \quad \frac{1}{1+\frac{1}{x}} \times \frac{-1}{x^2} dx = dt$
$\Rightarrow \quad \frac{1}{\frac{x+1}{x}} \times \frac{-1}{x^2} dx = dt$
$\Rightarrow \quad \frac{-x}{x^2(x+1)} dx = dt$
$\Rightarrow \quad \frac{d x}{x(x+1)}=- dt$
Putting $\log \left(1+\frac{1}{x}\right)= t$ and $\frac{d x}{x(x+1)}=- dt$ in equation (i), we get
$I=-\int t d t$
$=-\frac{t^2}{2}+c$
$=-\frac{1}{2}\left(\log \left(1+\frac{1}{x}\right)\right)^2+c$
$\therefore I=-\frac{1}{2}\left(\log \left(1+\frac{1}{x}\right)\right)^2+c$
Let $\log \left(1+\frac{1}{x}\right)= t$ then,
$d\left(\log \left(1+\frac{1}{x}\right)\right)= dt$
$\Rightarrow \quad \frac{1}{1+\frac{1}{x}} \times \frac{-1}{x^2} dx = dt$
$\Rightarrow \quad \frac{1}{\frac{x+1}{x}} \times \frac{-1}{x^2} dx = dt$
$\Rightarrow \quad \frac{-x}{x^2(x+1)} dx = dt$
$\Rightarrow \quad \frac{d x}{x(x+1)}=- dt$
Putting $\log \left(1+\frac{1}{x}\right)= t$ and $\frac{d x}{x(x+1)}=- dt$ in equation (i), we get
$I=-\int t d t$
$=-\frac{t^2}{2}+c$
$=-\frac{1}{2}\left(\log \left(1+\frac{1}{x}\right)\right)^2+c$
$\therefore I=-\frac{1}{2}\left(\log \left(1+\frac{1}{x}\right)\right)^2+c$
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