How to evaluate this indefinite integral?

[shakgoku]

Homework Statement

\int \frac{\sqrt{x^4-1}}{x^3}dx

Homework Equations

The Attempt at a Solution

tried substituting x^4 = \sec^2 \theta to get rid of square root but it was of no use because, I got another complex integral \int \frac{\sin^2\theta}{\cos^{\frac{3}{4}} \theta}d\theta

Is there an easy way to solve this problem?

 

[n.karthick]

why can't u take x^3 inside square root. Then it will be simplified to
sqrt(1-x^2)/x which is easier to integrate

 

[rn315656]

test!
The quadratic formula is
http://texify.com/$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

 

[shakgoku]

why can't u take x^3 inside square root. Then it will be simplified to
sqrt(1-x^2)/x which is easier to integrate

I don't get it. if we take x^3 inside we get,

\int \sqrt{\frac{x^4-1}{x^6}}dx=\int \sqrt{\frac{1}{x^2}-\frac{1}{x^6}}dx

 

[shakgoku]

I just realized I made a blunder in calculation.
That substitution yields
\int \frac{\sin^2\theta}{cos \theta}d \theta=\int \frac{1-\cos^2\theta}{\cos \theta} d\theta

and finally it is equal to,

\int \sec \theta d\theta-\int \cos\theta d\theta = \ln |x^2+\frac{\sqrt{1-x^2}}{x^2}|-\sqrt{1-x^2}+C

Which is the answer!