Just use trig sub.
$\displaystyle \int\frac{\sqrt{1-x^{2}}}{x^{2}}dx$
Let $\displaystyle x=sin{\theta}, \;\ dx=cos{\theta}d{\theta}$
Make the subs and it simplifies down to:
$\displaystyle \int{cot^{2}{\theta}}d{\theta}$
$\displaystyle \int{cot^{n}u}du=\frac{-1}{n-1}cot^{n-1}u-\int{cot^{n-2}u}du$
After you integrate cot, then make the sub $\displaystyle {\theta}=sin^{-1}(x)$ to get it back in terms of x.
You're finished.
simple answer: coz you did it wrong.
split it up first: ∫√[1/x^4 + 1/x²] dx
then do the substitution, u = 1/x, dx = -1/u² du
-∫√[u^4 + u²] * 1/u² du
factor out u from inside the square root and cancel one from denominator, you get
-∫√[u²-1]/u du
you can use a trig substitution to solve this.