Integral by parts instead of substitution

Question

Is it possible to integrate this by parts ? I know it's very simply to do it by substitution.

$$\int \frac{x}{\sqrt{1-x^2}}~dx$$

Answer 1

Yes, but it's pointless

$$\int \underbrace{x}_{u}\cdot \underbrace{\dfrac{1}{\sqrt{1-x^2}} dx}_{dv} = x\sin^{-1}(x) - \int \sin^{-1}(x) \ dx$$

The latter integral needs to be done by parts as well, unless you use a table, in which case you'll find

$$\int \sin^{-1}(x) \ dx = \sqrt{1-x^2} + x\sin^{-1}(x) + C$$

So you'll notice the $x\sin^{-1}(x)$ will cancel.