Simple double integral

[OrangeOne]

Hello,

I've tried solving the following double integral but cannot reach the key answer:

??1/((1+x+y)^2) dxdy
D

where D is a rectangle with the corners (0,0), (1,0), (1,2), (0,2)

This is the integral I set up:
? x goes from 0 to 1
? y goes from 0 to 2

??1/((1+x+y)^2) dxdy = ?? [(-1/(1+x+y))] = ?(-1/(x+3) + 1/(1+x)) dx= -ln(x+3) + ln (1+x)
(now I should put in the limits)

Have I done right so far?

The answer should be ln(3/2)

Can anyone please help? I would really appreciate it.

 

[galactus]

You have the correct limits and set up, but there is a problem in the execution.

\(\displaystyle \int_{0}^{2}\int_{0}^{1}\frac{1}{(1+x+y)^{2}}dxdy\)

\(\displaystyle \int\frac{1}{(1+x+y)^{2}}dx=\frac{-1}{1+x+y}\)

\(\displaystyle \frac{-1}{1+x+y}\Large{|}_{0}^{1}=\frac{1}{y+1}-\frac{1}{y+2}\)

\(\displaystyle \int_{0}^{2}\frac{1}{y+1}dy-\int_{0}^{2}\frac{1}{y+2}dy\)

\(\displaystyle [ln(y+1)-ln(y+2)]|_{0}^{2}=ln(3)-ln(2)=ln(\frac{3}{2})\)

 

[OrangeOne]

Now I understand, I didnt know ln(3) - ln(2) was ln(3/2)!

Thank you galactus!