# How do you evaluate the integral int 1/(x(1+(lnx)^2)?

Feb 5, 2017

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

#### Explanation:

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

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

This is the arctangent integral:

$= \arctan \left(u\right) + C = \arctan \left(\ln x\right) + C$