3
$\begingroup$

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.

  • 0
    @EnviousPage Good point! Changed it and clarified in the text too.2012-09-09

3 Answers 3

3

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.

1

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.

0

Wolfram alpha can solve this, but it is messy. Type

integral of ( (1 / (a + b*x + c*x^2) ) (1 / sqrt(d + e*x^2) ) )

  • 0
    I don't either, but some traditions are hard to break... Abramowitz & Stegun, Gradshteyn & Ryzhik are classics and easily fly (although they do have errors occasionally). Besides, Mathematica won't necessarily give you the simplest form of the expression... it's just what it managed to find via its CAS routines2012-09-09