Evaluate $\int \frac{ln x}{(1 + ln x)^2} \; dx $
Answer:
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 $
$= \int \frac{te^t}{(1+t)^2} \; dt = \frac{-te^t}{1 + t} + \int e^t \; dt$
$= \frac{-te^t}{1 + t} + e^t + C$
$= \frac{e^t}{1 + t} + C = \frac{x}{1 + ln x} + C$
