# Question #5266a

Nov 9, 2016

$- \frac{1}{3 {\left(\ln \left(x\right)\right)}^{3}} + C$

#### Explanation:

$I = \int \frac{1}{x {\left(\ln \left(x\right)\right)}^{4}} \mathrm{dx}$

We can use substitution $u = \ln \left(x\right)$. Differentiating this shows that $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x}$, or $\mathrm{du} = \frac{1}{x} \mathrm{dx}$. This is already present in our integral, but we can make it clearer:

$I = \int \frac{1}{\ln \left(x\right)} ^ 4 \left(\frac{1}{x} \mathrm{dx}\right) = \int {\underbrace{{\left(\ln \left(x\right)\right)}^{-} 4}}_{{u}^{-} 4} {\overbrace{\left(\frac{1}{x} \mathrm{dx}\right)}}^{\mathrm{du}} = \int {u}^{-} 4 \mathrm{du}$

We can now use the integration rule $\int {u}^{n} \mathrm{du} = {u}^{n + 1} / \left(n + 1\right) + C$:

$I = {u}^{- 4 + 1} / \left(- 4 + 1\right) + C = {u}^{- 3} / \left(- 3\right) + C = - \frac{1}{3 {u}^{3}} + C$

Since $u = \ln \left(x\right)$:

$I = - \frac{1}{3 {\left(\ln \left(x\right)\right)}^{3}} + C$