# How do you integrate h(t)=root3t(t^2+4) using the product rule?

Feb 3, 2017

The integral is $\left(\frac{3}{10}\right) {t}^{\frac{10}{3}} + \left(\frac{16}{3}\right) {t}^{\frac{4}{3}} + C$ which can be simplified to: $\left({t}^{\frac{2}{3}} / 10\right) \left(3 {t}^{5} + 16 {t}^{2}\right) + C$

#### Explanation:

The most important thing to understand is that there is no product rule or quotient rule for Integration. This is a mistake many students make. Sometimes, you can use u-substitution to deal with a problem like this, but first you should try something simpler. Let's use algebra to simplify it. Remember that $\sqrt[3]{t} = {t}^{\frac{1}{3}}$.

So, this is:
$\int \left({t}^{\frac{1}{3}} \left({t}^{2} + 4\right)\right)$

$\int \left({t}^{\frac{7}{3}} + 4 {t}^{\frac{1}{3}}\right)$ We now use the power rule:

${t}^{\frac{10}{3}} / \left(\frac{10}{3}\right) + \left(4 {t}^{\frac{4}{3}} / \left(\frac{4}{3}\right)\right) + C$ $=$

$\left(\frac{3}{10}\right) {t}^{\frac{10}{3}} + \left(\frac{16}{3}\right) {t}^{\frac{4}{3}} + C$

If you enjoy simplifying, you could do this:

$\left(\frac{9}{30}\right) {t}^{\frac{10}{3}} + \left(\frac{48}{30}\right) {t}^{\frac{4}{3}} + C$=

$\left(\frac{3}{30}\right) {t}^{\frac{2}{3}} \cdot \left(3 {t}^{5} + 16 {t}^{2}\right) + C$ = $\left({t}^{\frac{2}{3}} / 10\right) \left(3 {t}^{5} + 16 {t}^{2}\right) + C$