Integral of Sin-1 logx

Question

Sin^-1logx find integral 

Answer 1

This integral is not expressible in terms of standard functions.

 

Answer 2

From a Table of integrals I find ∫sin^-1u du = sin^-1u + √(1-u²) + C  so I get

 

antiderivative= ln x sin^-1(lnx) +  √(1-ln²x) + C.

 

Checking this answer by differentiating very carefully, applying both chain rule and product rule I DO get its derivative to be identical to the original integrand, so I think my answer is correct.