Solution:

\(\int\frac{\sqrt{1-x^2}}{x^4}dx\)

Let \(x=\sin\theta\Rightarrow dx=\cos\theta\cdot d\theta\)

\(=\int\frac{\sqrt{1-\sin^2\theta}}{\sin^4\theta}\cos\theta d\theta\)

\(=\int\frac{\cos^2\theta}{\sin^4\theta}d\theta=\int\cot^2\theta\cdot\csc^2\theta d\theta\)

Let \(u=\cot\theta\Rightarrow du=-\csc^2\theta d\theta\)

\(=-\int u^2 du\)

\(=-\frac{u^3}{3}+c\)

\(=-\frac{\cot^3\theta}{3}+c\)

\(=-\frac{(1-x^2)\sqrt{1-x^2}}{3\times3}+c\)

\(\int\frac{\sqrt{1-x^2}}{x^4}dx=-\frac{(1-x^2)\sqrt{1-x^2}}{3\times3}+c\)