Integral of $\int{\frac{dx}{(\arcsin{x})\sqrt{1-x^2}}}$

Do not use Integration by Parts. Use $u$-substitution. Let $u=\sin^{-1}(x)$. Then $du=dx/\sqrt{1-x^2}$

So now your integral is $$\int \frac{du}{u}$$

Note: Your comment at the end is incorrect. It is not true that $$\int\frac{dx}{\arcsin x} = \ln(\arcsin x)+ C\tag{Wrong!}$$

This is, unfortunately, a common mistake. Don't fall for it again!

While $$\int\frac{dx}{x} = \ln|x|+C$$ is true, in general, $$\int\frac{dx}{f(x)}$$ is not equal to $\ln|f(x)|+C$. If you differentiate $\ln|f(x)|$, you'll notice that you get $\frac{f'(x)}{f(x)}$. It is precisely because you have a $\frac{1}{\sqrt{1-x^2}}$ that you do get the natural logarithm of an arcsine.

"Another" method, which in fact is like substitution but, with some practice and care, perhaps is a little faster. Since $\,\displaystyle{(\arcsin x)'=\frac{1}{\sqrt{1-x^2}}}\,$ , and assuming we know $\,\displaystyle{\int\frac{f'}{f}\,dx=\log|f|+C}\,$, we can write

$$\int\frac{dx}{\sqrt{1-x^2}\arcsin x}\,dx=\int \frac{1}{\arcsin x}\,\frac{1}{\sqrt{1-x^2}}\,dx=$$$$=\int \frac{1}{\arcsin x}\,d(\arcsin x)=\log|\arcsin x|+C$$