Double Integral Plus Integration by Parts with Natural Log Problem

[M1ZeN]

Homework Statement

My homework problem is the double integral of y/1+xy dxdy. It is a definite double integral and both integrands have the values of a = 0 and b = 1.

Homework Equations

Integration by parts: uv - int(vdu)

The Attempt at a Solution

My first step of the double integral is I set:

u = 1+xy (with respect to x)
du = y

Then that gave me the integral of int(du/u) which equaled to ln(1+xy) ] b=1, a=0

I plug in the integrand values which gives me:

int[ln(1+y)]dy

Now this is where I'm having trouble. I do recognize this becomes integration by parts. So this is what I did:

u = ln(1+y) v = y
du = 1/(1+y) dv = dy

= y*ln(1+y) - int[y/(1+y)]dy

Then I set:

u = y v = ln(1+y)
du = dy dv = 1/(1+y)

= y*ln(1+y) - {y*ln(1+y) - int[ln(1+y)dy]}

***This is where I'm stumped

Appreciate any feedback! :)

 

[Dick]

You are going in circles with the integration by parts. To integrate y/(1+y) first use polynomial division to write it as 1-1/(1+y).