$\int \frac{\mathrm{dx}}{{\mathrm{x}}^2{\left({\mathrm{x}}^4+1\right)}^{3/4}}=\mathrm{A}{\left(\frac{{\mathrm{x}}^4+1}{{\mathrm{x}}^4}\right)}^{\mathrm{B}}+\mathrm{C}$

Solution:

$\mathrm{I}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2{\left({\mathrm{x}}^4+1\right)}^{3/4}}=\int \frac{\mathrm{dx}}{{\mathrm{x}}^2\cdot {\mathrm{x}}^3{\left(1+\frac{1}{{\mathrm{x}}^4}\right)}^{3/4}}$

$\begin{array}{r}\mathrm{Put}1+{\mathrm{x}}^{-4}=\mathrm{t}\Rightarrow \frac{-4}{{\mathrm{x}}^5}\mathrm{dx}=\mathrm{dt}\Rightarrow -\frac{1}{4}\int \frac{\mathrm{dt}}{{\mathrm{t}}^{3/4}}\\ =\frac{1}{4}\int {\mathrm{t}}^{-3/4}\mathrm{dt}=-\frac{1}{4}\cdot \frac{{\mathrm{t}}^{1/4}}{1/4}+\mathrm{C}\\ =-{\left(1+\frac{1}{{\mathrm{x}}^4}\right)}^{1/4}+\mathrm{C}=-{\left(\frac{1+{\mathrm{x}}^4}{{\mathrm{x}}^4}\right)}^{1/4}+\mathrm{C}\end{array}$

$\text{ Hence, }\mathrm{A}=-1,\mathrm{B}=\frac{1}{4}$