# How to evaluate $\int_{-\infty}^{\infty}(1+kx^2)^{-2}dx$

Can someone give a hint to evaluate the following integral? $$\int_{-\infty}^{\infty}(1+kx^2)^{-2}dx$$ where $k>0$.

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By taking $x=\frac{1}{\sqrt{k}}\tan(\theta)$ so, you have $$\int\frac{dx}{(1+kx^2)^2}\longrightarrow\int\frac{dt}{(1+\tan^2(t))\sqrt{k}}$$ which is elementary.

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+1 for (the method and) the exact amount of information needed to lead the OP to a full solution. –  Did Oct 29 '12 at 7:48
@did: The exact amount needed depends on the OP. It may well have been better to give a hint that leads the OP to figure out the substitution for himself, rather than just tell him what it is. –  Hurkyl Oct 29 '12 at 7:56
Thanks for the answer. Can you just explain me how did you guess the transformation $x=\frac{1}{\sqrt{k}}\tan(\theta)$? –  Kumara Oct 29 '12 at 9:10
@Kumara: The intuitive idea that I took that substitution was: the statements like $(1+x^2)$ may be convert to a simpler version if one took $x=\tan(t)$. So for example; $(1+x^2)^2$ would be $(1+\tan^2(t))^2=\cos^{-4}(t)$. And of course you know why we choose a coefficient constant $\frac{1}{\sqrt{k}}$. This is a standard method but not the only one as Hurkyl (+1) suggested. –  Babak S. Oct 29 '12 at 11:36
@Kumara: Quadratic polynomials appearing in integrands are very often simplified by completing the square (not needed here) and then making an appropriate trigonometric substitution so that you can use the pythagorean identities to simplify it further. You've probably seen integrals like $$\int \sqrt{1 + x^2} \, dx$$ and this is the same idea. –  Hurkyl Oct 29 '12 at 18:14

Hint:

Try integrating by parts integral $$\int\frac{dx}{1+kx^2}$$

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By integration by parts, I see $\int_{-\infty}^{\infty}\frac{dx}{1+kx^2}dx=2k \int_{-\infty}^{\infty}\left(\frac{x}{1+kx^2}\right)^2dx$. After that... –  Kumara Oct 29 '12 at 9:13

Partial fractions would work, if you're comfortable with complex numbers.

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Can you elaborate a bit more? –  Kumara Oct 29 '12 at 9:16
@Kumara: It's hard to guess what more you want to hear, since the complete algorithm for partial fractions should be in your book. I'm guessing it's the appearance of complex numbers: the roots of $1 + kx^2$ are complex, so when you factor it into linears, complex numbers will appear. (Of course, if the method of partial fractions reminds you to treat quadratics with imaginary roots by trigonometric substitution, that's just as good) –  Hurkyl Oct 29 '12 at 18:18