# How do you integrate -1 / (x(ln x)^2)?

Jun 11, 2016

$\frac{1}{\ln} x + C$

#### Explanation:

We have:

$\int - \frac{1}{x {\left(\ln x\right)}^{2}} \mathrm{dx} = - \int {\left(\ln x\right)}^{-} \frac{2}{x} \mathrm{dx}$

We can use substitution here, since the derivative of $\ln x$, which is $\frac{1}{x}$, is present alongside $\ln x$.

Let $u = \ln x$ such that $\mathrm{du} = \frac{1}{x} \mathrm{dx}$.

We then have:

$- \int {\left(\ln x\right)}^{-} \frac{2}{x} \mathrm{dx} = - \int {\left(\ln x\right)}^{-} 2 \left(\frac{1}{x}\right) \mathrm{dx} = - \int {u}^{-} 2 \mathrm{du}$

Integrate this with the rule: $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$, where $n \ne 1$.

Thus,

$- \int {u}^{-} 2 \mathrm{du} = - \left({u}^{- 2 + 1} / \left(- 2 + 1\right)\right) + C = - \left({u}^{-} \frac{1}{- 1}\right) + C = \frac{1}{u} + C$

Since $u = \ln x$,

$= \frac{1}{\ln} x + C$