# How do you find the integral int sqrt(x^3+x^2)(3x^2+2x)dx using substitution?

Mar 26, 2018

Let $u = {x}^{3} + {x}^{2}$. Then $\mathrm{du} = 3 {x}^{2} + 2 x \mathrm{dx}$ and $\mathrm{dx} = \frac{\mathrm{du}}{3 {x}^{2} + 2 x}$.

$I = \int \sqrt{u} \frac{3 {x}^{2} + 2 x}{3 {x}^{2} + 2 x} \mathrm{du}$

$I = \int \sqrt{u} \mathrm{du}$

$I = \frac{2}{3} {u}^{\frac{3}{2}} + C$

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

Hopefully this helps!

Mar 26, 2018

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

#### Explanation:

Here,

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

Substituting

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

$\implies I = \int \sqrt{{u}^{2}} 2 u \mathrm{du}$

$= \int u 2 u \mathrm{du}$

$= 2 \int {u}^{2} \mathrm{du}$

$= 2 \left[{u}^{3} / 3\right] + c$

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

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