Practice: Indefinite with Inverse Trig

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Compute the following integral: arcsinx1x2dx\displaystyle \int\frac{\text{arcsin}{x}}{\sqrt{1-x^2}}\,\text{d}x
(a) =[arctan(x)]22+C=\frac{[\arctan(x)]^2}{2}+C (b) =[arcsin(x)]22+C=\frac{[\arcsin(x)]^2}{2}+C (c) =121x2+C=\frac12\sqrt{1-x^2}+C (d) diverges


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