# How do you integrate int sqrt(x^3+x^2)(3x^2+2x) using substitution?

Mar 3, 2018

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

#### Explanation:

Let's set $u = {x}^{3} + {x}^{2.}$

Differentiating, we get

$\frac{\mathrm{du}}{\mathrm{dx}} = 3 {x}^{2} + 2 x$

We can change this derivative to a differential by moving $\mathrm{dx}$ to the right side (multiplying both sides by it)

$\mathrm{du} = \left(3 {x}^{2} + 2 x\right) \mathrm{dx}$

Look at our integral, $\int \sqrt{{x}^{3} + {x}^{2}} \left(3 {x}^{2} + 2 x\right) \mathrm{dx}$

We see $\left(3 {x}^{2} + 2 x\right) \mathrm{dx}$ appears in the integral; thus, we can replace it with $\mathrm{du} .$ Furthermore, we already said $u = {x}^{3} + {x}^{2} ,$ so we can replace everything under the root with $u .$

$\int \sqrt{{x}^{3} + {x}^{2}} \left(3 {x}^{2} + 2 x\right) \mathrm{dx} = \int \sqrt{u} \mathrm{du}$

Let's rewrite this with exponents:

$\int \sqrt{u} \mathrm{du} = \int {u}^{\frac{1}{2}} \mathrm{du}$

Recall that $\int {x}^{a} \mathrm{dx} = {x}^{a + 1} / \left(a + 1\right) + C$; where $C$ is just the constant of integration; therefore,

$\int {u}^{\frac{1}{2}} \mathrm{du} = {u}^{\frac{1}{2} + 1} / \left(\frac{1}{2} + 1\right) + C = {u}^{\frac{3}{2}} / \left(\frac{3}{2}\right) + C = \frac{2}{3} {u}^{\frac{3}{2}} + C$

Let's rewrite in terms of $x ,$ recalling that $u = {x}^{3} + {x}^{2} :$

$\frac{2}{3} {u}^{\frac{3}{2}} + C = \frac{2}{3} {\left({x}^{3} + {x}^{2}\right)}^{\frac{3}{2}} + C$