What I tried :
$$\int \frac {\log x}{\sqrt {1-x^2}} dx$$
$$= \log x \int \frac {1}{\sqrt {1-x^2}} dx - \int \frac{1}{x} \left(\int \frac {1}{\sqrt {1-x^2}} \, dx\right) \, dx$$
Now,
$$\int \frac {1}{\sqrt {1-x^2}} \, dx = \arcsin x + c$$
But inputting this into the $\displaystyle\int \frac{1}{x} \int \left(\frac {1}{\sqrt {1-x^2}} \, dx\right) \,dx$ part is not getting me into a neat solution. Looks like there is some complex solution of the problem, but I don't know how to solve complex integration. What I did is :
$$\int \frac{1}{x} \int \left(\frac {1}{\sqrt {1-x^2}} \, dx\right) \, dx$$
$$= \int y \cot y \, dy \text{ where } y = \arcsin x$$
$$= y \ln |\sin y| - \int \ln |\sin y |dy$$
So will you plz help me in one of the two ways:
$\displaystyle\int y \cot y \, dy = \text{?}$ or $\displaystyle\int \ln |\sin y | \, dy = \text{?}$
Any better/neater solution of the original problem ?
I saw this question in MSE but in definite integral form, (which is okay), but not sure about this indefinite form.Thanks.