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

by dave_western
Tags: integrate, y2dy
 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: 3,348 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 $$dv = dy$$