Evaluate $\int \frac{\sqrt{1+x^8} dx}{x^{13}}$ 
Evaluate: $\displaystyle\int \frac{\sqrt{1+x^8}}{x^{13}}dx$

My attempt:
I have tried substituting $1+x^8=z^2$ but that did not work. I also tried writing $x^{13}$ in the denominator as $x^{16}.x^{-3}$ hoping that it would bring the integrand into some form but that too did not work.
 A: The substitution $x=\tan^{1/4}t$ gives $$\int\frac{\sqrt{1+x^8}}{x^{13}}dx=\int\frac{1}{4}\sin^{-4}t\cos tdt=-\frac{1}{12}\sin^{-3}t+C=-\frac{1}{12}\bigg(\frac{x^8}{1+x^8}\bigg)^{-3/2}+C.$$
A: Let $x^4=\tan u$ so that $dx=\frac{\sec^2 u}{4x^3}du$
so our integral becomes
$\int \frac{\sqrt{1+\tan^2 u}}{x^{13}}.\frac{\sec^2 u}{4x^3}du$
or rather
$\frac{1}{4}\int \frac{\sec^3 u}{\tan^4 u}du$
which is
$\frac{1}{4}\int \cos u\sin^{-4} u du$
which equals
$-\frac{1}{12}\sin^{-3}u +C$
now substitute back from $u$ to $x$ which I think is
$-\frac{1}{12}(\frac{x^8}{1+x^8})^\frac{-3}{2}+C$
A: $x=\dfrac1y,dx=-\dfrac{dy}{y^2}$
$$\int\dfrac{\sqrt{1+x^8}}{x^{13}}dx=-\int\dfrac{y^{13}\sqrt{y^8+1}}{y^4\cdot y^2}dy$$
Set $\sqrt{y^8+1}=u $ or $y^8+1=v$
A: Write the integral as $$\int \frac {\sqrt {1+\frac {1}{x^8}}}{x^9} dx$$ Use the substitution $$1+\frac {1}{x^8}=t^2$$ to get the new integral as $$\int \frac {-t^2}{4}dt$$ Which is very elementary to do and then after obtaining the answer resubstitute the value of $t$.
A: Hint We can rewrite the radical in the integrand in the familiar form $\sqrt{1 + u^2}$ if we substitute $$u = x^4, \qquad du = 4 x^3 \,dx .$$
