# How do you find the integral of  x^3/(1+x^2)?

Oct 10, 2015

$\frac{1}{2} \left(1 + {x}^{2} - \ln \left(1 + {x}^{2}\right)\right) + C$

#### Explanation:

We'll do integration by substitution. Take:
$u = {x}^{2}$
$\mathrm{du} = 2 x$
Substituting this in, we get:

$\frac{1}{2} \setminus \int \frac{u}{1 + u} \mathrm{du}$
Now take:
$t = 1 + u$
$\mathrm{dt} = \mathrm{du}$

Substitute this in:
$\frac{1}{2} \setminus \int \frac{t - 1}{t} \mathrm{dt} = \frac{1}{2} \setminus \int 1 - \frac{1}{t} \mathrm{dt} = \frac{1}{2} \left(\setminus \int 1 \mathrm{dt} - \setminus \int \frac{1}{t} \mathrm{dt}\right)$
$= \frac{1}{2} \left(t - \ln \left(t\right)\right) + C$
Now we need to change back to $x$:
$= \frac{1}{2} \left(1 + u - \ln \left(1 + u\right)\right) + C$
$= \frac{1}{2} \left(1 + {x}^{2} - \ln \left(1 + {x}^{2}\right)\right) + C$