5.5 The Substitution Rule/41

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

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

<math> \begin{align}

\int \frac{1}{u} du = \ln |u| +c = \ln |\arcsin {x}| + c