Integrating $\int \frac{dx}{x^{11}\sqrt{1+x^4}}$

Find $$I=\int \frac{dx}{x^{11}\sqrt{1+x^4}}$$

With $x^2=t$, we get $I=\int _{ }^{ }\frac{d\ \sqrt{t}}{\left(\sqrt{t}\right)^{11}.\sqrt{1+t^2}}=\frac{1}{2}\int _{ }^{ }\frac{dt}{t^6\sqrt{1+t^2}}$

Now, with $t= \tan k$ we get $I=\frac{1}{2}\int _{ }^{ }\frac{\cos ^5k.\ dk}{\sin ^6k}$ .

At this step, I can not do anything more. Can you show me the way to solve this?

I try to go with this way but it doesn't work.

Solution 1:

After substituting $t = \frac{1}{x^4}$:

$$I = -\frac{1}{4} \int \frac{x^5 \ dt}{x^{11} \sqrt{1+x^4}} = -\frac{1}{4} \int \frac{t^2 x^2 \ dt}{\sqrt{x^4(t+1)}}= -\frac{1}{4}\int \frac{t^2 \ dt}{\sqrt{1+t}}.$$

Now further substitute $u = 1 + t$ and expand $t^2 = (u - 1)^2$.