The value of \(\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}}\) is

Let \(\begin{aligned} \mathrm{I} & =\int \frac{\mathrm{d} x}{x^2\left(x^4+1\right)^{\frac{3}{4}}} \\ & =\int \frac{\mathrm{d} x}{x^2\left[x^4\left(1+\frac{1}{x^4}\right)\right]^{\frac{3}{4}}} \\ & =\int \frac{\mathrm{d} x}{x^5\left(1+\frac{1}{x^4}\right)^{\frac{3}{4}}} \\ & =\int \frac{\left(1+\frac{1}{x^4}\right)^{\frac{-3}{4}}}{x^5}\end{aligned}\)
\(\begin{aligned} & \quad \text { Let } 1+\frac{1}{x^4}=\mathrm{t} \\ & \Rightarrow \frac{-4}{x^5} \mathrm{~d} x=\mathrm{dt} \\ & \Rightarrow \frac{\mathrm{d} x}{x^5}=-\frac{\mathrm{dt}}{4} \\ & \therefore \quad \mathrm{I}=\frac{1}{4} \int \frac{\mathrm{t}^{\frac{-3}{4}}}{\frac{-3}{4}} \mathrm{dt} \\ & \quad=\frac{-1}{4} \frac{t^{\frac{-3}{4}}+1}{\frac{-3}{4}}+1 \\ & \quad=-t^{\frac{1}{4}}+\mathrm{c}\end{aligned}\)
\(\begin{aligned} \Rightarrow I & =-\left(1+\frac{1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c} \\ & =-\left(\frac{x^4+1}{x^4}\right)^{\frac{1}{4}}+\mathrm{c} \\ \Rightarrow \mathrm{I} & =\frac{-\left(x^4+1\right)^{\frac{1}{4}}}{x}+\mathrm{c}\end{aligned}\)