# Homework Help: Help with an intermediate integral

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1. Dec 31, 2014

### FallenLeibniz

1. The problem statement, all variables and given/known data
I have been trying to evaluate an integral that has come up in the process of me solving a different problem, but am completely stuck. As I have confirmed with Wolfram Alpha that the integral once solved yields the correct solution to my problem. However, I am trying to better my ability to solve integrals without the help of tables or computer aids, so I am trying to find the correct solution to this one. Also, because my issue deals more with the math rather than the physics of the problem, I have included it here among the Math section. If this was inappropriate to put this here or if this is too little to prove that I have done enough work, please let me know and I will take this post down and include the full problem in the Physics section.

The integral is $\int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz$

The problem here simply is that every time I attempt this integral, the math becomes a "runaway train". I can use any help or hints with regards to how to fully evaluate this integral.

2. Relevant equations

$\int_{-a}^{a}\frac{1}{(z^2+a^2)(z^2+2a^2)^{1/2}} dz$

3. The attempt at a solution

I started off by using integration by parts and designating $dv=[z^2+a^2]^{-1}$ and $u=[z^2+2a^2]^{-1/2}$. This led to the result of $\frac{1}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-1/2}\bigg|_{-a}^a +\int_{-a}^a\frac{z}{a}(arctan(\frac{z}{a}))[z^2+2a^2]^{-3/2} dz$. When I try to evaluate the integral in this difference, that's when I go into a long never ending train of Integration by Parts. Any help with this integration would be must appreciated.

2. Dec 31, 2014

### Ray Vickson

Preliminary simplification is always worthwhile. If you call your integral $F(a)$, changing variables to $z = ay$ gives
$$F(a) = \frac{1}{a^2} F(1), \\ \text{where } \;\; F(1) = \int_{-1}^1 \frac{dy}{(y^2+1)\sqrt{y^2+2}} = 2 \int_0^1 \frac{dy}{(y^2+1)\sqrt{y^2+2}}$$
Submit this last integration to Wolfram Alpha (a free on-line version of Mathematical lite) to see what you get.

Alternatively, try the change of variable $u = y/\sqrt{y^2+2}$.