I did the integration by parts and got this expression, but then I am stuck on how to take it further. I tried substituting u=1-x^2, but then I had to do a partial fraction decomposition (which I did not take). Any hints or help would be very appreciated!
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$\begingroup$Let $\sqrt{1-x^2}=t$.
Thus, $-\frac{2x}{2\sqrt{1-x^2}}dx=dt$ and $$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int\left(\frac{1}{x\sqrt{1-x^2}}\cdot\frac{\sqrt{1-x^2}}{x}\right)dt=$$ $$=-\int\frac{1}{x^2}dt=-\int\frac{1}{1-t^2}dt=\frac{1}{2}\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt=$$ $$=\frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+C=\frac{1}{2}\ln\left|\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}+1}\right|+C.$$
$\endgroup$ $\begingroup$Take $x = \sin u$ initially to get simple integral.
$$\int u \csc( u )\cot( u) du= -u\csc( u) + \int \csc (u) du\\ = -u \csc (u) +\ln|\cot( u )-\csc (u)| + c.$$
$\endgroup$ $\begingroup$I wrote my answer for this question, click here to see the image.
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