5.5 The Substitution Rule/41

$\int {\frac {1}{{\sqrt {1-x^{2}}}\arcsin {x}}}=\int {\frac {1}{u}}du=\ln |u|+c=\ln |\arcsin {x}|+c$

${\begin{aligned}u&=\arcsin {x}\\[2ex]du&={\frac {1}{\sqrt {1-x^{2}}}}dx\\[2ex]\end{aligned}}$

${\begin{aligned}\int {\frac {1}{u}}du&=\ln |u|+c\\[2ex]&=\ln |\arcsin {x}|+c\\[2ex]\end{aligned}}$