Question

How do you find the Integral of `ln(2x+1)`?

Answer 1

It is `int(ln(2x+1))dx=xln(2x+1)-x+1/2ln(2x+1)`

Explanation:

We will use integration by parts

`intln(2x+1)dx=int(x)'ln(2x+1)dx=xln(2x+1)-intx*(ln(2x+1))'dx=xln(2x+1)-int(x*2/(2x+1))=xln(2x+1)-int(2x+1)/(2x+1)-1/(2x+1)dx=xln(2x+1)-x+1/2ln(2x+1)`