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
 A: If you want to proceed using integration by parts, then we have the following
$$\begin{align}
I&=\int \frac{dx}{x(\log x)^2}\\\\
&=\frac{1}{\log x}+2\int \frac{dx}{x(\log x)^2}-C\\\\
&=\frac{1}{\log x}+2I-C
\end{align}$$
whereupon solving for $I$ reveals that 
$$I=-\frac{1}{\log x}+C$$
A: $$\begin{align}
\int\frac1{x\ln^2x}\,dx & = \int u^{-2}\,du,\quad\text{Let }u=\ln x,du=\frac{dx}x\\
& = -u^{-1}+C\\
& = -\frac1{\ln x}+C.
\end{align}$$
A: 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 $\frac{x^2}{4}$.)
