Evaluate the definite integral please.

Question

anonymous · Answer

$$\int\limits_{0}^{1} \frac{logx}{\sqrt{1-x^2}} dx$$

anonymous · Answer

Wow. That is a doozy. Try plugging it into WA, it gives back a crazy indefinite integral that I wouldn't know how to solve even with the limits given.

anonymous · Answer

:'(

psymon · Answer

That's some nasty substitution if it exists O.o

anonymous · Answer

The problem is correctly typed? Looking at the other questions you've asked, they're difficult but this one isn't likely to be solved with pen and paper.

zzr0ck3r · Answer

lol

anonymous · Answer

Wow. Not sure how. But wow.

anonymous · Answer

$$\int_{0}^1\frac{\ln(x)}{\sqrt{1-x^2}} dx =-\frac{\pi}{2}\ln(2)$$

zzr0ck3r · Answer

we can all type the integral into a calculator, how did you get the answer is the goal...

anonymous · Answer

It can be evaluated in terms of several trigonometric functions, some logarithms, and a dilogarithm, you need to know the value of a special function, though I am sure there are other ways.

anonymous · Answer

If he comes back with all the steps I'm going to order a truck full of humble pies.

anonymous · Answer

I am to lazy to do the work, why don't you post this on mse and wait

zzr0ck3r · Answer

lol @amishjeb

loser66 · Answer

I think it is improper integral since the upper limit is 1 and it makes the integral undefined. therefore we have to use limit of improper integral there 
$$lim_{t\rightarrow 1}\int_0^t {\frac{logx}{\sqrt{1-x^2}}}dx$$