Explanation
Concept:
∫xn dx=n+1xn+1+c
Calculation:
I = ∫x32x4−2x2+1x2−1 dx
=41∫x3x4(2−x22+x41)4x2−4 dx
=41∫x52−x22+x414x2−4 dx
\begin{aligned}
=\frac{1}{4}\int\frac{\frac{4}{x^{3}}-\frac{4}{x^{5}}}{\sqrt{\left(2-\frac{2}{x^{2}}+\frac{1}{x^{4}}\right)}}\mathrm{dx} \\<br> \mathrm{Let2-\frac{2}{x^{2}}+\frac{1}{x^{4}}=t} \\
\text{Differentiating with respect to x, we get} \\
\left({\frac{4}{x^{3}}}-{\frac{4}{x^{5}}}\right)\mathrm{dx}=\mathrm{dt} \\<br> \mathrm{Now}, \\<br> \mathbf{I}={\frac{1}{4}}\int{\frac{\mathrm{dt}}{\sqrt{t}}} \\
=\frac{1}{4}\int\mathrm{t}^{-1/2}\mathrm{dt} \\<br> =\frac{1}{4}\times\frac{\mathrm{t}^{\frac{1}{2}}}{\frac{1}{2}}+\mathrm{c} \\
=\frac{1}{2}\mathrm{t}^{\frac{1}{2}}+\mathrm{c} \\<br> =\frac{1}{2}\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}}+\mathrm{c} \\
=\frac{\sqrt{2\mathrm{x}^{4}-2\mathrm{x}^{2}+1}}{2\mathrm{x}^{2}}+\mathrm{C}
\end{aligned}
