3

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.

3 Answers3

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.

Gerry Myerson
  • 179,216
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.

Sasha
  • 70,631
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) ) )

binn
  • 1,291
  • Yes, I have the result from W|A and Mathematica too, but I need either an outline to do it or an explicit reference (e.g. standard, published integral tables) so that I can cite it in a publication. "Integrated using W|A" won't fly in a review... –  Sep 09 '12 at 01:49
  • @fsk: I have seen "integrated using mathematica" fly in a review. – binn Sep 09 '12 at 01:54
  • Yes, but it depends on the rigor of the journal... it works in a soft science/biology journal, but not so in an applied physics journal. –  Sep 09 '12 at 01:56
  • I don't see how it's different than using an integral table. – binn Sep 09 '12 at 01:59
  • 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 routines –  Sep 09 '12 at 02:06