
#1
Jan1607, 04:05 PM

P: 5

Stuck at the end of a double integral, still have to integrate ln(4+y^2)dy
Assuming I did the right first step. Original double integral is x/(x^2+y^2) Thanks! 



#2
Jan1607, 09:21 PM

HW Helper
P: 1,007

What are the bounds on x and y?




#3
Jan1607, 09:27 PM

P: 5

the region R = [1,2] * [0,1]




#4
Jan1607, 09:39 PM

Emeritus
Sci Advisor
PF Gold
P: 16,101

Integrate ln(4+y^2)dy?
Have you had any thoughts on integrating that? I see two obvious things to try:
(1) Do what you normally do with integrals of logarithms. (2) Make a substitution. 



#6
Jan1707, 01:06 AM

Sci Advisor
HW Helper
P: 11,863

There's no need for substitution for that integral. Part integration once then a smart move in the numerator of the remaining integral and it's done.
Daniel. 



#7
Jan1707, 03:39 AM

P: 361

is the integral we are talking about [tex]\int_{0}^{1}\int_{1}^{2}\frac{x}{x^2+y^2}dxdy[/tex]?
i get [tex]\frac{1}{2}\int_{0}^{1}\left[\ln(4+y^2)\ln(1+y^2)\right]dy[/tex] what should i do next? (edit: i got it. integration by parts.) 



#8
Jan1707, 06:34 PM

HW Helper
P: 3,353

Murshid, split the integral up, do them separately perhaps? Eg, say for [tex]\int ln(1+y^2) dy[/tex] we could let y^2 equal u. Find du/dy, easy. Then, solve for dy. Now substitute that value in. We end up with [tex]\int \frac{ln (u+1)}{2(u)^{1/2}} du [/tex]. Then some nice integration by parts and we are done?
Takes a while though, I hope your patient. 



#9
Jan1807, 01:56 AM

P: 361

well i got it already. thanks anyway.
but we can directly use integration by parts on this [tex]\int ln(1+y^2) dy[/tex] by letting [tex]u = \ln(1+y^2)[/tex] and [tex]dv = dy[/tex] 


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