<s>[INST] $\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$ 
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!
 [/INST] $\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx\;,$ Now Let 
Let $\displaystyle x = \frac{1}{t}\;,$ Then $\displaystyle dx = -\frac{1}{t^2}dt\;,$ Then Integral Convert into 
$\displaystyle = -\int \frac{t^3}{(1+t^4)^{\frac{3}{4}}}dt\;,$ Now Let $(1+t^4) = u\;, $ Then $\displaystyle t^3dt = \frac{1}{4}du$
So Integral $\displaystyle = -\frac{1}{4}\int t^{-\frac{3}{4}}dt = -u^{\frac{1}{4}}+\mathcal{C} = -\left(1+t^4\right)^{\frac{1}{4}}+\mathcal{C}$
So Integral $\displaystyle \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx = - \left(\frac{1+x^4}{x^4}\right)^{\frac{1}{4}}+\mathcal{C.}$
</s>