# How do you find the definite integral of  (xdx) / ((x^2 + 1)ln(x^2 + 1)) from [1,2]?

Aug 24, 2016

$.191$

#### Explanation:

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

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

$\int \frac{1}{u} \frac{1}{2} \mathrm{du}$

$\frac{1}{2} \left[\ln u\right]$

now replace the u in terms of x

$\frac{1}{2} {\left[\ln \left(\ln \left({x}^{2} + 1\right)\right)\right]}_{1}^{2}$

$\frac{1}{2} \left[\ln \left(\ln \left({2}^{2} + 1\right)\right) - \ln \left(\ln \left({1}^{2} + 1\right)\right)\right] = .191$