Solving with integration by parts: $\int \frac 1 {x\ln^2x}dx$ 
Solving: $$\int \frac 1 {x\ln^2x}dx$$ with parts.

$$\int \frac 1 {x\ln^2x}dx= \int \frac {(\ln x)'} {\ln^2x}dx \overset{parts} = \frac {1} {\ln x}-\int \frac {(\ln x)} {(\ln^2x)'}dx$$
$$\int \frac {(\ln x)} {(\ln^2x)'}dx=\int \frac {(x\ln x)} {(2\ln x)}dx=\frac {x^2} 4$$
So $$\int \frac 1 {x\ln^2x}dx= \frac {1} {\ln x} -  \frac {x^2} 4$$
But the answer should be $-\frac {1} {\ln x}$ and I can't find which step is wrong...
http://www.integral-calculator.com/#expr=1%2F%28x%28lnx%29%5E2%29

As pointed out in previous answers, this can be found most easily using the substitution $u=\ln x$.
Using integration by parts, though, with $u=(\ln x)^{-2}$ and $dv=\frac{1}{x}dx$,  so $du=-2(\ln x)^{-3}dx$ and $v=\ln x$,
gives $\displaystyle\int\frac{1}{x(\ln x)^2}dx=(\ln x)^{-1}-(-2)\int(\ln x)^{-2}\frac{1}{x}dx=(\ln x)^{-1}+2\int\frac{1}{x(\ln x)^2}dx$, so
$\displaystyle-\int\frac{1}{x(\ln x)^2}dx=(\ln x)^{-1}+C\;\;$ and $\;\;\displaystyle\int\frac{1}{x(\ln x)^2}dx=-(\ln x)^{-1}+C$

(Notice that you have a mistake in the 3rd line of your answer where you get $\frac{x^2}{4}$.)
