# How do I integrate this (or is there a solution in a table)?

How do I integrate:

$$\int \frac{1}{p+q(x-r)^2}\frac{1}{\sqrt{s+t x^2}}\, dx$$

All variables other than $x$ can be assumed to be greater than $0$ and independent of $x$. Pointers to a formula from an integration table are also sufficient. This wikipedia article is almost what I need, except for the fact that I have an $x-r$ in just one of the terms.

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What is y? Is it a function of x? Or a constant? –  Martin Leslie Sep 9 '12 at 1:39
@Martin Everything other than $x$ is independent of $x$. I've changed it to $e$ to avoid confusion :) –  user39836 Sep 9 '12 at 1:40
Using $e$ will confuse people thinking it's the natural log base e. A variable like $m$ or $n$ would probably be better. –  Envious Page Sep 9 '12 at 1:42
@EnviousPage Good point! Changed it and clarified in the text too. –  user39836 Sep 9 '12 at 1:44

This may be the hard way, but

1. Substitute $x=\sqrt{s/t}\tan\theta$. That will get rid of the square root, and turn the integrand into a rational function of trig functions.

2. Use the $\tan(t/2)$ substitution to turn the integrand into a rational function.

3. Use partial fractions to do the resulting problem.

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As an alternative to Gerry's suggestion, you could Euler's substitution. Let's relabel $t$ to $t^2$, i.e. we are solving for $$\int \frac{1}{p+q(x-r)^2} \frac{1}{\sqrt{s + t^2 x^2}} \mathrm{d} x$$

Specifically, make a change of variables $$x = \frac{u^2-s}{2 t u}, \quad \mathrm{d}x = \frac{u^2+s}{2 t u^2} \mathrm{d} u, \quad \frac{1}{\sqrt{s + t^2 x^2}} = \frac{2 u}{s+u^2}, \quad \frac{\mathrm{d}x}{\sqrt{s + t^2 x^2}} =\frac{1}{t}\frac{\mathrm{d}u}{u}$$ Thus: $$\int \frac{1}{p+q(x-r)^2} \frac{1}{\sqrt{s + t^2 x^2}} \mathrm{d} x = \int \frac{1}{p+ q \left(\frac{u^2-s}{2 t u}-r\right)^2} \frac{1}{t} \frac{\mathrm{d}u}{u} = \int \frac{4 t u \cdot \mathrm{d} u}{4 p t^2 u^2 + q \left(u^2 -s - 2 r t u\right)^2}$$ Now it is down to integration of the rational function.

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