## Is there a simpler way to integrate this?

 Quote by aruwin I calculated another 2 times but I didn't get the answer :( Are you sure my working there had only one mistake?
If you fix up changing the - (-2 + z) to + (-2 + z) and fix up your mistake with the 2^7 and make it 2*(-1)^7 and plug everything in correctly you will get 104/5. Show us your working so we can see what you did.

 Quote by chiro If you fix up changing the - (-2 + z) to + (-2 + z) and fix up your mistake with the 2^7 and make it 2*(-1)^7 and plug everything in correctly you will get 104/5. Show us your working so we can see what you did.
Oh,thank you!!!!I got it,I got it!!!
By the way, I have another question on multiple integration but must I create another thread or can I just continue in here?

 Quote by aruwin Oh,thank you!!!!I got it,I got it!!! By the way, I have another question on multiple integration but must I create another thread or can I just continue in here?
Just post it in here, it's on topic and a simple continuation of the original threads premise.

 Quote by chiro Just post it in here, it's on topic and a simple continuation of the original threads premise.
This seems a little more complicated because there's fraction which is hard to decompose and I don't know how to decompose it anyway. Can you help me out with this one?

J = ∫∫D [(x-y)/(x+y)^3] dxdy

D = {(x,y) | 0 ≤x≤1, 0≤y≤1}

Recognitions:
Homework Help
 Quote by aruwin This seems a little more complicated because there's fraction which is hard to decompose and I don't know how to decompose it anyway. Can you help me out with this one? J = ∫∫D [(x-y)/(x+y)^3] dxdy D = {(x,y) | 0 ≤x≤1, 0≤y≤1}
Try the substitution u = x - y, v = x+y. Remember to work out the Jacobian and the new limits.

 Quote by Mute Try the substitution u = x - y, v = x+y. Remember to work out the Jacobian and the new limits.
I did it the other way,though. But I didn't get the answer that I am supposed to. Can you check for me,please?

 Recognitions: Homework Help The integral of y/(1+y)2 is not just -y/(1+y). You missed a term. By the way, this is actually a pretty insidious integral. After you add the term you missed and get the result, try doing the integral in the opposite order (y first, then x). My earlier suggestion to try a substitution might not be a good alternate way to do this in light of this result.

 Quote by Mute The integral of y/(1+y)2 is not just -y/(1+y). You missed a term. By the way, this is actually a pretty insidious integral. After you add the term you missed and get the result, try doing the integral in the opposite order (y first, then x). My earlier suggestion to try a substitution might not be a good alternate way to do this in light of this result.
Yes, I got it! The integral that I got is -1/2 but how come the answer that my professor gave is ∞? Is there any explanation to that?

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 Quote by aruwin Yes, I got it! The integral that I got is -1/2 but how come the answer that my professor gave is ∞? Is there any explanation to that?
Did you try doing the integration in the opposite order yet? (y first, then x). I suspect that your professor was not referring to the integral you just performed, but rather the integral of the absolute value of the integrand. To deduce that the integral of |(x-y)/(x+y)3| is infinity, you should do the integral you just solved in the order I suggested and understand what that result means.