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

  • Thread starter dave_western
  • Start date
  • Tags
    Integrate
In summary, after discussing different approaches, it was concluded that the given double integral of x/(x^2+y^2) can be solved by using integration by parts. It was suggested to split the integral and then use integration by parts on ln(1+y^2) dy. Alternatively, a substitution could also be used.
  • #1
dave_western
5
0
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!
 
Physics news on Phys.org
  • #3
the region R = [1,2] * [0,1]
 
  • #4
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.
 
  • #5
eg. Integration by parts then trig sub.
 
  • #6
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
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:
  • #8
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
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]
 

1) What is the formula for integrating ln(4+y^2)dy?

The formula for integrating ln(4+y^2)dy is ∫ln(4+y^2)dy = yln(4+y^2) - 2y + C.

2) Can we simplify ln(4+y^2)dy before integrating?

Yes, we can use logarithmic properties to simplify ln(4+y^2)dy to 2ln(2+y)dy, which makes it easier to integrate.

3) Is it possible to use substitution to solve this integral?

Yes, we can use substitution by letting u = 4+y^2, du = 2ydy, and substituting them into the formula for integrating ln(4+y^2)dy.

4) Does the constant C have any significance in the integration of ln(4+y^2)dy?

Yes, the constant C represents the constant of integration and is added to account for all possible solutions to the integral. It is important to include the constant when solving indefinite integrals.

5) Are there any special techniques required to integrate ln(4+y^2)dy?

No, there are no special techniques required to integrate ln(4+y^2)dy. However, it may be helpful to use logarithmic properties or substitution to simplify the integral before integrating.

Similar threads

  • Calculus
Replies
11
Views
2K
Replies
2
Views
2K
Replies
3
Views
216
Replies
1
Views
812
Replies
2
Views
180
  • Calculus
Replies
4
Views
1K
  • Calculus
Replies
24
Views
3K
Replies
8
Views
955
Replies
5
Views
1K
Back
Top