# How to compute $\int \sqrt{\frac{x^2-3}{x^{11}}}dx$

How to compute the following integral ??

$\displaystyle\int \sqrt{\frac{x^2-3}{x^{11}}}dx$

I use the wolfram Alpha to get an ugly answer

http://www.wolframalpha.com/input/?i=integral++%28%28x^2-3%29%2Fx^%2811%29%29^0.5+dx

But there is no step for me

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If you are looking for a simple expression it does not exists. You integral has the form $$\int (x^2-3)^{\frac{1}{2}}x^{-\frac{11}{2}} dx$$ The expression of the form $$\int (a+bx^n)^px^m dx$$ is called a differential binomial and can be integrated in elementary functions only if $p$ is an integer (not), $(m+1)/n$ is an integer (not), or $p+(m+1)/n$ (not).

Added You can use the substitutions $x=t^\alpha$, where $\alpha$ is the common denominator of $m$ and $n$; $a+bx^n=t^{\alpha}$ where $\alpha$ is the denominator of $p$; $ax^{-n}+b=t^\alpha$, where $\alpha$ is the denominator of $p$ correspondingly.

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thanks for you help – cwk709394 Dec 29 '12 at 17:27
@BabakSorouh Thanks, I added the substitutions that work in each case. – Artem Dec 29 '12 at 17:51

The WolframAlpha answer suggests using a substitution of the form $x = \sqrt{3} \sin^2 u$, but I'd suggest using $x = \sqrt{3} \csc^2 u$, if you want to avoid complex-valued results. You'll get an integrand with a factor of $\sqrt{1+\sin^2u}$ in it. To get an answer, you're going to have to integrate by parts (and use an identity) a few times to reduce the order/degree, until you get down to integrands of the form $1/\sqrt{1+\sin^2u}$ and $\sqrt{1+\sin^2u}$, whose antiderivatives are elliptic functions of the first kind and the second kind, $F(u\,|\,{-1})$ and $E(u\,|\,{-1})$, respectively. If you have the patience to figure that all out, you'll realize the WA answer is not uglier than one should expect, except perhaps that it gives complex values instead of real ones.

(You can also integrate your original integral by parts first, using the identity $-3 = (x^2-3)+x^2$, until you reach an integrand of the form $x^{1/2}\sqrt{x^2-3}$; and then do the trig. sub.)

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