# How do you integrate int t^2(t^3+4)^(-1/2) using substitution?

Apr 3, 2018

The integral is $\frac{2}{3} \sqrt{{t}^{3} + 4} + C$

#### Explanation:

First, rewrite the integral:

$\textcolor{w h i t e}{=} \int {t}^{2} {\left({t}^{3} + 4\right)}^{- \frac{1}{2}} \mathrm{dt}$

$= \int {t}^{2} / {\left({t}^{3} + 4\right)}^{\frac{1}{2}} \mathrm{dt}$

$= \int {t}^{2} / \sqrt{{t}^{3} + 4} \mathrm{dt}$

Now, let:

$u = {t}^{3} + 4 \quad \textcolor{b l u e}{\implies} \quad \mathrm{du} = 3 {t}^{2} \mathrm{dt} \quad \textcolor{b l u e}{\implies} \quad \mathrm{dt} = \frac{\mathrm{du}}{3 {t}^{2}}$

Substituting:

$= \int {t}^{2} / \sqrt{u} \cdot \frac{\mathrm{du}}{3 {t}^{2}}$

$= \int \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{t}^{2}}}}}{\sqrt{u}} \cdot \frac{\mathrm{du}}{3 \textcolor{r e D}{\cancel{\textcolor{b l a c k}{{t}^{2}}}}}$

$= \int \frac{1}{\sqrt{u}} \cdot \frac{\mathrm{du}}{3}$

$= \frac{1}{3} \int \frac{1}{\sqrt{u}} \mathrm{du}$

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

$= \frac{1}{3} \int {u}^{- \frac{1}{2}} \mathrm{du}$

$= \frac{1}{3} \cdot {u}^{- \frac{1}{2} + 1} / \left(- \frac{1}{2} + 1\right)$

$= \frac{1}{3} \cdot {u}^{\frac{1}{2}} / \left(\frac{1}{2}\right)$

$= \frac{1}{3} \cdot 2 \cdot \sqrt{u}$

$= \frac{2}{3} \sqrt{u}$

Put ${t}^{3} + 4$ back in for $u$ (and don't forget to add $C$):

$= \frac{2}{3} \sqrt{{t}^{3} + 4} + C$

That's it. Hope this helped!