How do I integrate ln(4+y^2)dy at the end of a double integral?

[dave_western]

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!

 

[benorin]

What are the bounds on x and y?

 

[dave_western]

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

 

[Hurkyl]

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.

 

[mjsd]

eg. Integration by parts then trig sub.

 

[dextercioby]

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.

 

[murshid_islam]

is the integral we are talking about $$\int_{0}^{1}\int_{1}^{2}\frac{x}{x^2+y^2}dxdy$$?

i get $$\frac{1}{2}\int_{0}^{1}\left[\ln(4+y^2)-\ln(1+y^2)\right]dy$$
what should i do next? (edit: i got it. integration by parts.)

 

[Gib Z]

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

Takes a while though, I hope your patient.

 

[murshid_islam]

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