# Integrate ln(4+y^2)dy?

by dave_western
Tags: integrate, y2dy
 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!
 HW Helper P: 1,007 What are the bounds on x and y?
 P: 5 the region R = [1,2] * [0,1]
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P: 16,094

## 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.
 HW Helper P: 858 eg. Integration by parts then trig sub.
 HW Helper Sci Advisor P: 11,716 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.
 P: 361 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.)
 HW Helper P: 3,353 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.
 P: 361 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 $$dv = dy$$

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