# How do you integrate x^3 sqrt(x^2 + 1) dx?

Jan 26, 2017

$\int {x}^{3} \sqrt{1 + {x}^{2}} \mathrm{dx} = \frac{1}{15} \left(3 {x}^{2} - 2\right) \left(1 + {x}^{2}\right) \sqrt{1 + {x}^{2}} + C$

#### Explanation:

Substitute $t = 1 + {x}^{2}$, $\mathrm{dt} = 2 x \mathrm{dx}$:

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

Now solve the integral in $t$:

$\frac{1}{2} \int \left(t - 1\right) \sqrt{t} \mathrm{dt} = \frac{1}{2} \int t \sqrt{t} \mathrm{dt} - \frac{1}{2} \int \sqrt{t} \mathrm{dt} = \frac{1}{2} \int {t}^{\frac{3}{2}} \mathrm{dt} - \frac{1}{2} \int {t}^{\frac{1}{2}} \mathrm{dt} = \frac{1}{2} {t}^{\frac{5}{2}} / \left(\frac{5}{2}\right) - \frac{1}{2} {t}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C = \frac{1}{5} {t}^{\frac{5}{2}} - \frac{1}{3} {t}^{\frac{3}{2}} + C = \frac{1}{15} {t}^{\frac{3}{2}} \left(3 t - 5\right) + C$

Substituting back $x$:

$\int {x}^{3} \sqrt{1 + {x}^{2}} \mathrm{dx} = \frac{1}{15} {\left(1 + {x}^{2}\right)}^{\frac{3}{2}} \left(3 \left(1 + {x}^{2}\right) - 5\right) + C$

$\int {x}^{3} \sqrt{1 + {x}^{2}} \mathrm{dx} = \frac{1}{15} \left(1 + {x}^{2}\right) \sqrt{1 + {x}^{2}} \left(3 {x}^{2} - 2\right) + C$