Question

Please let me know if I'm on the right track.
I'm trying to solve the integral x/(1-x^2)^(1/2) dx
I have u=x du=dx dv=1/(1-x^2)^(1/2)
v=2(1-x^2)^(1/2)?

Answer 1

what is v ?
you don't need any v here, just a u:
\[u=1-x^2\]\[du=-2xdx\to xdx=-\frac{1}{2}du\]\[\int\frac{x}{(1-x^2)^{1/2}}dx=-\frac{1}{2}\int\frac{du}{u^{1/2}}=-\frac{1}{2}\int u^{-1/2}du\]integrate then sub back in at the end for u.