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 $dg=\frac{1}{(1+u)^2}du.$
Set $t=\ln x$ and $dt=\frac{dx}{x}$

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

By parts $f=e^tt$ and $dg=\frac{dt}{(1+t)^2}$

$$=-\frac{e^tt}{t+1}+\int e^t dt=\color{red}{\frac{x}{\ln x+1}+\mathcal C}$$