Integrate ln(4+y^2)dy?

Main Question or Discussion Point

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!
 

Answers and Replies

benorin
Homework Helper
1,067
14
What are the bounds on x and y?
 
the region R = [1,2] * [0,1]
 
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,843
17
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
Homework Helper
725
3
eg. Integration by parts then trig sub.
 
dextercioby
Science Advisor
Homework Helper
Insights Author
12,965
536
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.
 
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.)
 
Last edited:
Gib Z
Homework Helper
3,344
4
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.
 
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]
 

Related Threads for: Integrate ln(4+y^2)dy?

  • Last Post
Replies
4
Views
8K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
1K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
18K
  • Last Post
Replies
4
Views
2K
Replies
26
Views
4K
  • Last Post
Replies
5
Views
9K
Top