<s>[INST] Evaluation of $\int\frac{1}{x^2.(x^4+1)^{\frac{3}{4}}}dx$ 
Evaluate the integral
  $$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx$$

My Attempt:
Let $x = \frac{1}{t}$. Then $dx = -\frac{1}{t^2}\,dt$. Then the integral converts to 
$$
-\int \frac{t^3}{(1+t^4)^{3/4}}\,dt
$$
Now Let $(1+t^4) = u$. Then $t^3\,dt = \frac{1}{4}du$. This changes the integral to
$$
\begin{align}
-\frac{1}{4}\int t^{-3/4}\,dt &= -u^{1/4}+\mathcal{C}\\
&= -\left(1+t^4\right)^{1/4}+\mathcal{C}
\end{align}
$$
So we arrive at the solution
$$\int\frac{1}{x^2\left(x^4+1\right)^{3/4}}\,dx = - \left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C.}$$
Question: Is there any other method for solving this problem?
 [/INST] Alternate Solution:
$$
\begin{align}
\int\frac{1}{x^2\cdot \left(x^4+1\right)^{3/4}}\,dx &= \int\frac{1}{x^2\cdot x^3\cdot \left(1+x^{-4}\right)^{3/4}}\,dx\\
&= \int\frac{1}{x^{5}\cdot \left(1+x^{-4}\right)^{3/4}}\,dx
\end{align}
$$
Now Let $(1+x^{-4}) = t$. Then $x^{-5}\,dx = -\frac{1}{4}\,dt$, so we have
$$
-\frac{1}{4}\int \frac{1}{t^{3/4}}dt = -\frac{1}{4}\cdot 4\cdot t^{1/4}+\mathcal{C} = -\left(\frac{1+x^4}{x^4}\right)^{1/4}+\mathcal{C}
$$ 
</s>