What is #int (lnsqrt(3x+2))^2dx#?

1 Answer
Nov 2, 2015

Here is some help to get you started.

Explanation:

#int (lnsqrt(3x+2))^2dx#

Let's look at the integrand for a bit:

# (lnsqrt(3x+2))^2#

First note that # (lnsqrt(3x+2)) = ln(3x+2)^(1/2) = 1/2ln(3x+2)#

and #3x+2# is linear, so we can do a substitution.

Then the question will be how to find

#int (lnt)^2 dt#

Details

#int(lnsqrt(3x+2))^2dx = int (1/2ln(3x+2))^2 dx#

# = 1/4 int (ln(3x+2))^2 dx#

With # t = 3x+2#, the integral becomes:

# = 1/12 int (lnt)^2 dt#

Now try substitution, try parts, try anything else you think of.

Hint
Recall that we can integrate #lnx# by using integration by parts.