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!
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  $$ \begin{align} -\frac{1}{4}\int t^{-3/4}\,dt &amp;= -u^{1/4}+\mathcal{C}\\ &amp;= -\left(1+t^4\right)^{1/4}+\mathcal{C} \end{align} $$  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?
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