Integral of $\int{\frac{dx}{(\arcsin{x})\sqrt{1-x^2}}}$ I am having a problem solving an integral. I am stuck in an infinite loop. Integral is:
$$\int{\frac{dx}{\sqrt{1-x^2}\arcsin{x}}}$$
I have separated it in dv and u on this way:
$$u = \frac{dx}{\sqrt{1-x^2}}$$
$$dv = \frac{1}{\arcsin{x}}$$
And the using:
$$u v - \int{v \, du}$$
I get again:
$$\int{\frac{dx}{\sqrt{1-x^2}\arcsin{x}}}$$
I dont know, but probably, I am doing something wrong. I am new at solving Integrals so I am learning :) According to my book the result should be:
$$\ln({\arcsin{x}})-C$$
And it will be true if I didn't had $$\sqrt{1-x^2}$$ but on this way I have no idea.
 A: 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.
A: "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
$$\int\frac{dx}{\sqrt{1-x^2}\arcsin x}\,dx=\int \frac{1}{\arcsin x}\,\frac{1}{\sqrt{1-x^2}}\,dx=$$$$=\int \frac{1}{\arcsin x}\,d(\arcsin x)=\log|\arcsin x|+C$$
A: 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 $$\int \frac{du}{u}$$
A: $\arcsin x = u$ then $\frac{1}{\sqrt{1-x^2}}dx = du $. In the integral it will then be: $ \int u^{-1}du  $ which is $\ln \vert u \vert +c$ or simply $\ln \vert \arcsin x\vert+c$.
