Integrating $\frac{\ln x}{(1+\ln x)^2}$ 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...
 A: 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 $dg=\frac{1}{(1+u)^2}du.$
A: 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.
A: Set $t=\ln x$ and $dt=\frac{dx}{x}$
$$=\int\frac{e^tt}{(t+1)^2}dt$$
By parts $f=e^tt$ and $dg=\frac{dt}{(1+t)^2}$
$$=-\frac{e^tt}{t+1}+\int e^t dt=\color{red}{\frac{x}{\ln x+1}+\mathcal C}$$
A: $1+\ln x=u\iff x=e^{u-1},dx=e^{u-1}du$
$$I=\int\dfrac{\ln x}{(1+\ln x)^2}dx=\int\dfrac{u-1}{u^2}e^{u-1}du=\int e^{u-1}\left(\dfrac1u+\dfrac{d(1/u)}{du}\right)du =\dfrac{e^{u-1}}u+K$$
as $\dfrac{e^xf(x)}{dx}=e^x[f(x)+f'(x)]\implies\int e^x[f(x)+f'(x)]dx=e^x f(x)$
$$\implies I=\dfrac x{1+\ln x}+K$$
Alternatively, 
$$\dfrac{\ln x}{(1+\ln x)^2}=\dfrac{1+\ln x-1}{(1+\ln x)^2}$$
$$=\dfrac1{1+\ln x}\cdot\dfrac{dx}{dx}+x\cdot\dfrac{d\{1/(1+\ln x)\}}{dx}$$
What is $\displaystyle\dfrac{d(uv)}{dx}=?\implies\int(u\dfrac{dv}{dx}+v\dfrac{du}{dx})dx=uv+K$
