My first integration problem

$u=\arcsin(x)$ $\sin u = x$ Substituting that back in you get $\displaystyle \int \frac{\text{d}u}{1-\sin^{2} u}$ .

That's the same as $\displaystyle \int \frac{\text{d}u}{\cos^{2} u}$ or $\displaystyle \int \sec^{2}u \text{ d} u$ .

That gives $\tan u + C$ . But I wanted to get rid of $u$ . So I substitute back $u=\arcsin(x)$ . $= \tan\left( \arcsin x \right) + C$ But that isn't an answer choice, so I had to simplify further. $= \frac{\sin \left( \arcsin x \right)}{\cos \left( \arcsin x \right)}+ C$ The numerator is just $x$ . For the denominator, we know that $\cos(x) = \sqrt{1-\sin^{2}x}$ $= \frac{x}{\sqrt{1 - \sin^{2} \left( \arcsin x \right)}}+ C$ $=\boxed{ \displaystyle \frac{x}{\sqrt{1-x^{2}}} + C}$