# How do you integrate int t(2t+7)^(1/3) by integration by parts method?

Jan 5, 2018

$\frac{3 t}{8} {\left(2 t + 7\right)}^{\frac{4}{3}} - \frac{9}{112} {\left(2 t + 7\right)}^{\frac{7}{3}} + C$

#### Explanation:

$\int t {\left(2 t + 7\right)}^{\frac{1}{3}} \mathrm{dt}$

Integration by parts tells us that:

$\int u v ' = u v - \int u ' v$

So we first need to determine which of our products is $u$ and which is $v '$. As a general rule of thumb, $u$ is usually the product that will get 'simpler' upon differentiating.

-So we will take $u$ and its respective derivative to be:

u = t; u'=1

-$v '$ and its respective integral to be:

v' = (2t+7)^(1/3); v=3/8(2t+7)^(4/3)

So the integral will now become:

$\int t {\left(2 t + 7\right)}^{\frac{1}{3}} \mathrm{dt} = \frac{3 t}{8} {\left(2 t + 7\right)}^{\frac{4}{3}} - \int \frac{3}{8} {\left(2 t + 7\right)}^{\frac{4}{3}} \mathrm{dt}$

We can now directly integrate that to finish the task and get:

$\frac{3 t}{8} {\left(2 t + 7\right)}^{\frac{4}{3}} - \frac{9}{112} {\left(2 t + 7\right)}^{\frac{7}{3}} + C$