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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.2012-10-29
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    @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.2012-10-29
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    @did: Thanks. Indeed, I am trying to be like you, Hurkyl and other masters here. :)2012-10-29
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    Thanks for the answer. Can you just explain me how did you guess the transformation $x=\frac{1}{\sqrt{k}}\tan(\theta)$?2012-10-29
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    @Kumara: In fact if you put that substitution, then the whole fraction would be turn into a simple trigonometric simple version. Can you do that?2012-10-29
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    @BabakSorouh Yes, I did that and the integral becomes $\int_{-\pi/2}^{\pi/2}\frac{d(\tan t)}{(1+\tan^2(t))\sqrt{k}}=\pi/\sqrt{k}$. But my question was, how did you guess that substitution. Is it a standard one or by a trial and error?2012-10-29
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    @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.2012-10-29
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    @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.2012-10-29
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    @Hurkyl The amount of information in Babak's answer seems to have been pretty close to *exact*, according to the comments above, don't you think?2012-10-29
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    @did: I thought exactly the opposite. Upon reflection, I think that we are focusing on different aspects. You're focusing on the fact that Babak gave Kumara the answer for the part Kumara didn't get. I'm focusing on the fact that Kumara wants to learn how to solve the part he didn't understand for himself.2012-10-29
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    (Just so I'm not misunderstood, I'm not criticizing Babak's answer. It's clear and concise, and surely a very good hint for a number of people who would have difficulty with this problem. In fact, (+1) now that I've thought about it)2012-10-29
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    @Hurkyl Babak did not *give Kumara the answer* since Kumara had to ask for some explanations. To my eyes, the Babak-Kumara exchange about Babak's answer looks very much like somebody *learning* some maths from somebody else. If only we could witness more of this on MSE... :-)2012-10-29
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    @did: The answer *+comment* combination works very well, I agree with that.2012-10-29
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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...2012-10-29
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Partial fractions would work, if you're comfortable with complex numbers.

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    Can you elaborate a bit more?2012-10-29
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    @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)2012-10-29