# int x/(1+x)(1+x^2)dx ?

May 21, 2018

$\int \frac{x}{1 + x} \left(1 + {x}^{2}\right) \mathrm{dx} = {x}^{3} / 3 - {x}^{2} / 2 + 2 x - 2 \ln \left(x + 1\right) + C$, $C \in \mathbb{R}$

#### Explanation:

$\int \frac{x}{1 + x} \left(1 + {x}^{2}\right) \mathrm{dx} = \int \frac{x}{1 + x} \mathrm{dx} + \int {x}^{3} / \left(1 + x\right) \mathrm{dx}$
let
${I}_{1} = \int \frac{x}{1 + x} \mathrm{dx} = \int \frac{1 + x - 1}{1 + x} \mathrm{dx} = \int \left(1 - \frac{1}{1 + x}\right) \mathrm{dx}$
$= \int 1 \mathrm{dx} - \int \frac{1}{1 + x} \mathrm{dx} = x - \ln \left(1 + x\right)$

${I}_{2} = \int {x}^{3} / \left(1 + x\right) \mathrm{dx} = \int \frac{{x}^{3} + 1 - 1}{1 + x} \mathrm{dx}$

$= \int \frac{{x}^{3} + 1}{x + 1} \mathrm{dx} - \int \frac{1}{1 + x} \mathrm{dx}$

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

$= {x}^{3} / 3 - {x}^{2} / 2 + x - \ln \left(x + 1\right)$

Now,

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

${I}_{1} = x - \ln \left(x + 1\right)$

${I}_{2} = {x}^{3} / 3 - {x}^{2} / 2 + x - \ln \left(x + 1\right)$

Thus,

$\int \frac{x}{1 + x} \left(1 + {x}^{2}\right) \mathrm{dx} = x - \ln \left(x + 1\right) + {x}^{3} / 3 - {x}^{2} / 2 + x - \ln \left(x + 1\right)$

Rearranging

$\int \frac{x}{1 + x} \left(1 + {x}^{2}\right) \mathrm{dx} = {x}^{3} / 3 - {x}^{2} / 2 + 2 x - 2 \ln \left(x + 1\right) + C$, $C \in \mathbb{R}$