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: 
**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.
$\arcsin x = u$ 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 $\ln \vert \arcsin x\vert+c$.