Question: 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.

Answer: 
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 $$\int \frac{du}{u}$$
$\arcsin x = u$ 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 $\ln \vert \arcsin x\vert+c$.