Evaluate, on an interval where the integrand is real, \[ \int\frac{x}{\sqrt{4-x^4}}\,dx. \]
Solution
Let \(u=x^2/2\), so \(du=x\,dx\) and \[ \sqrt{4-x^4}=2\sqrt{1-u^2}. \] Therefore \[ \int\frac{x}{\sqrt{4-x^4}}\,dx =\frac12\int\frac{du}{\sqrt{1-u^2}} =\boxed{\frac12\arcsin\!\left(\frac{x^2}{2}\right)+C.} \]