### A problem asked by Arpit Agarwal

Evaluate $\int \frac{ln x}{(1 + ln x)^2} \; dx$

Substitute $t = ln x$ and we have $dt = \frac{1}{x} \; dx$ or, $dx = x \; dt = e^t \; dt$ and

$\int \frac{ln x}{(1 + ln x)^2} \; dx = \int \frac{te^t}{(1+t)^2} \; dt$

Now substitute $u = te^t$ and $dv = \frac{1}{(1+t)^2} \; dt$, so we have $du = e^t(t + 1)$ and $v = \frac{-1}{1 + t}$

and we have

$\int u \; dv = uv - \int v \; du$

$= \int \frac{te^t}{(1+t)^2} \; dt = \frac{-te^t}{1 + t} + \int e^t \; dt$

$= \frac{-te^t}{1 + t} + e^t + C$

$= \frac{e^t}{1 + t} + C = \frac{x}{1 + ln x} + C$