Question: I came across a question today:

> 
> $$\int \frac{\ln x \, dx}{(1+\ln x)^2}$$
> 
> 
> 

How to do it? I tried to substitute $1+\ln x$ and also tried taking $\ln x$ outside the brackets but they doesn't work... I even tried to use that multiplication rule.That didn't work either. I can't see anything to pass through it...

Answer: 
Let $u=\ln x$ and $du=\frac{1}{x}dx$.

Thus, the integral transforms to $\displaystyle\int \dfrac{\ln x dx}{(1+\ln x)^2}=\displaystyle \int \frac{e^u u du}{(1+u)^2}.$

Now, integrate by parts with $f=e^u u$ and $dg=\frac{1}{(1+u)^2}du.$
**Hint:**
Wolfram Alpha solves this (in an admittingly cumbersome way) by first letting $v=\log x$ and then integrating by parts with $u = ve^v$ and $dg = \frac{dv}{(v+1)^2}$. Then just back substitute and you'll get your answer.