Question

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

Answer 1

`(3t)/8(2t+7)^(4/3)-9/112(2t+7)^(7/3)+C`

Explanation:

We start with:

`intt(2t+7)^(1/3)dt`

Integration by parts tells us that:

`intuv' = uv-intu'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:

`intt(2t+7)^(1/3)dt = (3t)/8(2t+7)^(4/3)- int3/8(2t+7)^(4/3)dt`

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

`(3t)/8(2t+7)^(4/3)-9/112(2t+7)^(7/3)+C`