$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$ 
Evaluate $$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$

I thought of rewriting this as $$\large{\int\dfrac{dx}{x^5(1+\frac{1}{x^4})^{3/4}}}$$ and substituting $$u^4=\left(1+\frac{1}{x^4}\right)\Rightarrow u=\left(1+\frac{1}{x^4}\right)^{1/4}$$ and subsequently I got $$du=\dfrac{1}{4}\left(1+\frac{1}{x^4}\right)^{-3/4}\times (-4x^{-5})dx$$
However, I can not think of how to proceed further. Any help would be truly appreciated. Many thanks in advance!

$$\int\frac{1}{x^5}\frac{1}{(1+\frac{1}{x^4})^{3/4}}dx$$
$$u=1+\frac{1}{x^4}$$
$$-\frac{1}{4}du=\frac{1}{x^5}$$
The integral in the variable $u$ is then
$$-\frac{1}{4}\int u^{-3/4}du$$
