This is a nasty one. I'd suggest starting with trig substitution. Are you familiar with this method. I'm posting Mathematica's solution.
Here's a website I found with some example of trig substitution. Draw a triangle, with one leg x^2, the other leg 1, and the hypotenuse as
http://tutorial.math.lamar.edu/AllBr...stitutions.asp
Let me name both integrals
I'll integrate one term, the other one goes completely the same. I'll do I.
As I said, we'll be getting an ln and an inverse tangent. To get the ln part, we find the derivative of the denominator.
We manipulate the nominator to get this derivative and split the integral.
So we have
Now the first integral has exactly the derivative of the denominator in the nominator, which gives the natural logarithm of the denominator as primitive function.
I'll continue with the second integral, which will give an inverse tangent as I said. We'll complete the square in the denominator.
But for an inverse tangent, we need something of the form (*)²+1, so a +1 instead of +1/2. We'll fix that by bringing the factor 1/2 for the integral and correcting the square.
We're almost there, we just have to integrate with respect to instead of x. I'll adjust the dx and correct before the integral.
Now it's a standard integral with the inverse tangent as primitive function.
Combining these
Completely analogous (only the signs change), we find for J
So we have found
I don't usually do such long solutions, but I found this a very good example to illustrate this technique which is often used for these kind of integrals. Most of it can be done slightly differently by using substitutions, but I found this a nice way to illustrate how it can be done by only manipulating the integrals.