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

1 Answer
Andrew I.
Jan 5, 2018

# (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#

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