$\int\dfrac{dx}{x^2(x^4+1)^{3/4}}$ 
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!
 A: We have $$\int\frac{1}{x^{2}\left(x^{4}+1\right)^{3/4}}dx\overset{u=1/x^{4}}{=}-\frac{1}{4}\int\frac{1}{\left(u+1\right)^{3/4}}du\overset{u+1=v}{=}-\frac{1}{4}\int\frac{1}{v^{3/4}}d=
 $$ $$=-\sqrt[4]{v}+C=-\frac{\sqrt[4]{x^{4}+1}}{x}+C.
 $$
A: $$\int\frac{1}{x^5}\frac{1}{(1+\frac{1}{x^4})^{3/4}}dx$$
$$u=1+\frac{1}{x^4}$$
$$-\frac{1}{4}du=\frac{1}{x^5}$$
The integral in the variable $u$ is then
$$-\frac{1}{4}\int u^{-3/4}du$$
A: Just taking up where you left off:
$$
-\int du\, = -u+C = -(1+\frac{1}{x^4})^{\frac{1}{4}}+C.
$$
That should be it.
A: Let, $1+\frac{1}{x^4}=t \implies \frac{-4dx}{x^5}=dt$$$\large{\int\dfrac{dx}{x^2(x^4+1)^{3/4}}}$$ $$=\int \frac{dx}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}}$$ $$=\frac{-1}{4}\int\frac{dt}{\left(t\right)^{3/4}}$$ $$=\frac{-1}{4}\int (t)^{-3/4}dt$$ $$=\frac{-1}{4}\frac{t^{1/4}}{1/4}+C$$ $$=-\left(1+\frac{1}{x^4}\right)^{1/4}+C$$$$=-\frac{(1+x^4)^{1/4}}{x}+C$$
A: $\bf{My\; Solution::}$ 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 $\displaystyle \int\frac{1}{x^2\left(x^4+1\right)^{\frac{3}{4}}}dx = - \left(\frac{1+x^4}{x^4}\right)^{\frac{1}{4}}+\mathcal{C.}$
