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

Apr 25, 2018

The answer ${\left({x}^{2} + 1\right)}^{\frac{5}{2}} / 5 - {\left({x}^{2} + 1\right)}^{\frac{3}{2}} / 3 + c$

#### Explanation:

Show the steps below

$\int {x}^{3} \cdot {\left({x}^{2} + 1\right)}^{\frac{1}{2}} \cdot \mathrm{dx}$

suppose:
$u = {x}^{2} + 1$
$x = {\left(u - 1\right)}^{\frac{1}{2}}$
$\mathrm{du} = 2 x \cdot \mathrm{dx}$
$\mathrm{dx} = \frac{1}{2 x} \cdot \mathrm{du}$
the int become after suppose:

$\int {\left[\sqrt{u - 1}\right]}^{3} \cdot \sqrt{u} \cdot \frac{1}{2 \cdot \sqrt{u - 1}} \cdot \mathrm{du}$

$\frac{1}{2} \int {\left[\sqrt{u - 1}\right]}^{2} \cdot \sqrt{u} \cdot \mathrm{du}$

$\frac{1}{2} \int \left(u - 1\right) \cdot \sqrt{u} \cdot \mathrm{du} = \frac{1}{2} \int u \cdot \sqrt{u} - \sqrt{u} \cdot \mathrm{du} = \frac{1}{2} \int {u}^{\frac{3}{2}} - {u}^{\frac{1}{2}} \cdot \mathrm{du}$

$\frac{1}{2} \left[\frac{2}{5} \cdot {u}^{\frac{5}{2}} - \frac{2}{3} \cdot {u}^{\frac{3}{2}}\right] + c$

${u}^{\frac{5}{2}} / 5 - {u}^{\frac{3}{2}} / 3 + c$

${\left({x}^{2} + 1\right)}^{\frac{5}{2}} / 5 - {\left({x}^{2} + 1\right)}^{\frac{3}{2}} / 3 + c$