# How to prove $\int_{0}^{1} \frac{\arccos x}{\ln x}dx=-\sum_{k=0}^{\infty}\frac {(2k-1)!!\ln(2k+2)}{2^k k! (2k+1)}$

Question:-Prove that $$\int_{0}^{1} \frac{\arccos x}{\ln x}dx=\sum_{k=0}^{\infty}-\frac {(2k-1)!!\ln(2k+2)}{2^k k! (2k+1)}$$.

I have no idea how to prove this, If we use Taylor series of both $$\arccos x$$ and $$\ln x$$, then it is difficult to combine them. Also, Taylor series of $$\frac {\arccos x}{\ln x}$$ gives no help.

Can anybody help me to prove this?

We have : $$\arccos x=\frac{\pi}{2}-\arcsin x=\arcsin 1-\arcsin x$$ And using Taylor series of $$\arcsin$$ we get : $$\arccos x=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!(2n+1)}(1-x^{2n+1})$$ Then we get : $$\int_0^1\frac{\arccos x}{\ln x}=\sum_{n=0}^\infty\frac{(2n-1)!!}{(2n)!(2n+1)}\biggl(\int_0^1\frac{1-x^{2n+1}}{\ln x}\biggr)$$ Therefore we put $$x=e^{-t}$$ so we have By Frullani Theorem : $$\int_0^1\frac{1-x^{2n+1}}{\ln x}=\int_0^\infty\frac{e^{-t}-e^{-(2n+2)t}}{-t}=-\ln (2n+2)$$ Then we get the result.