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

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