Please help me on computing this integral. Thank you!

Question

Answer 1

Given that

First take only integration after that apply limits

For ln (3x+2) , u = 2+3x du = 3dx

For integral ln u by integrate by parts

Where f = log u                 dg = du

            df = 1/u                    g = u

The integral of 1 is u

Substitute back  u = 2+3x

Which is equal to

Now apply limits

Substract the possible terms

The value of log 0 is 1

Multiply with e/3

Therefore the required solution is

Comments

The required solution = [- e/3 + e/3 ln(e)]

= [- e/3 + e/3 * 1]   [ since ln(e) = 1]

= [- e/3 + e/3]

= 0.

The solution is 0.