1
$\begingroup$

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

3 Answers 3

5

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.

  • 0
    @did: The answer *+comment* combination works very well, I agree with that.2012-10-29
2

Hint:

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

  • 0
    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
1

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

  • 1
    @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