# Find int \ x^5/(1+x^2)^4 \ dx ?

Aug 4, 2017

$\int \setminus {x}^{5} / {\left(1 + {x}^{2}\right)}^{4} \setminus \mathrm{dx} = - \frac{3 {x}^{4} + 3 {x}^{2} + 1}{6 {\left(1 + {x}^{2}\right)}^{3}} + C$

#### Explanation:

We want to find:

$I = \int \setminus {x}^{5} / {\left(1 + {x}^{2}\right)}^{4} \setminus \mathrm{dx}$

Let us perform a substitution:

Let u=1+x^2 => (du)/dx=2x; \ \ \ x^2=u-1

We can rewrite the integral and substitute as follows

$I = \int \setminus \frac{1}{2} {\left({x}^{2}\right)}^{2} / {\left(1 + {x}^{2}\right)}^{4} \setminus \left(2 x\right) \setminus \mathrm{dx}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus {\left(u - 1\right)}^{2} / {\left(u\right)}^{4} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{{u}^{2} - 2 u + 1}{{u}^{4}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} \setminus \int \setminus \frac{1}{u} ^ 2 - \frac{2}{u} ^ 3 + \frac{1}{{u}^{4}} \setminus \mathrm{du}$

Which we can now integrate to get:

$I = \frac{1}{2} \left(- \frac{1}{u} + \frac{1}{u} ^ 2 - \frac{1}{3 {u}^{3}}\right) + C$
$\setminus \setminus = - \frac{1}{2} \left(\frac{1}{u} - \frac{1}{u} ^ 2 + \frac{1}{3 {u}^{3}}\right) + C$
$\setminus \setminus = - \frac{1}{2} \left(\frac{3 {u}^{2} - 3 u + 1}{3 {u}^{3}}\right) + C$
$\setminus \setminus = - \frac{1}{6} \left(\frac{3 {u}^{2} - 3 u + 1}{{u}^{3}}\right) + C$

And by restoring the substitution we get:

$I = - \frac{1}{6} \left(\frac{3 {\left(1 + {x}^{2}\right)}^{2} - 3 \left(1 + {x}^{2}\right) + 1}{{\left(1 + {x}^{2}\right)}^{3}}\right) + C$
$\setminus \setminus = - \frac{1}{6} \left(\frac{3 {x}^{4} + 6 {x}^{2} + 3 - 3 - 3 {x}^{2} + 1}{{\left(1 + {x}^{2}\right)}^{3}}\right) + C$
$\setminus \setminus = - \frac{1}{6} \left(\frac{3 {x}^{4} + 3 {x}^{2} + 1}{{\left(1 + {x}^{2}\right)}^{3}}\right) + C$
$\setminus \setminus = - \frac{1}{6} \left(\frac{3 {x}^{4} + 3 {x}^{2} + 1}{{\left(1 + {x}^{2}\right)}^{3}}\right) + C$