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

Dec 3, 2016

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

#### Explanation:

Substitute:

$x = \tan t$

$\mathrm{dx} = \frac{\mathrm{dt}}{\cos} ^ 2 t$

$\int {x}^{3} / \left(1 + {x}^{2}\right) \mathrm{dx} = \int {\tan}^{3} \frac{t}{1 + {\tan}^{2} t} \frac{\mathrm{dt}}{\cos} ^ 2 t$

Use the trigonometric identity:

$1 + {\tan}^{2} t = \frac{1}{\cos} ^ 2 t$

$\int {x}^{3} / \left(1 + {x}^{2}\right) \mathrm{dx} = \int {\tan}^{3} \frac{t}{\frac{1}{\cos} ^ 2 t} \frac{\mathrm{dt}}{\cos} ^ 2 t = \int {\tan}^{3} t \mathrm{dt}$

Using the same identity again:

$\int {\tan}^{3} t \mathrm{dt} = \int \tan t {\tan}^{2} t \mathrm{dt} = \int \tan t \left(\frac{1}{\cos} ^ 2 t - 1\right) \mathrm{dt} =$

$= \int \tan t d \left(\tan t\right) - \int \tan t \mathrm{dt} = \frac{1}{2} {\tan}^{2} t - \int \sin \frac{t}{\cos} t \mathrm{dt} =$

$= \frac{1}{2} {\tan}^{2} t + \int \frac{d \left(\cos t\right)}{\cos} t = \frac{1}{2} {\tan}^{2} t + \ln | \cos t |$

To substitute back $x$, we have that:

$\tan t = x$

$\cos t = \pm \sqrt{\frac{1}{1 + {\tan}^{2} t}} = \pm \frac{1}{\sqrt{1 + {x}^{2}}}$

So:

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