Question Number 30508 by abdo imad last updated on 22/Feb/18
$${find}\:{I}=\:\int\:\:{e}^{{arcsinx}} {dx}\:. \\ $$
Commented by sma3l2996 last updated on 24/Feb/18
$${I}=\int{e}^{{arcsinx}} {dx} \\ $$$${u}={e}^{{arcsinx}} \Rightarrow{u}'=\frac{{e}^{{arcsinx}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${v}'=\mathrm{1}\Rightarrow{v}={x} \\ $$$${I}={xe}^{{arcsinx}} −\int\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{e}^{{arcsinx}} {dx}+{c}_{\mathrm{1}} \\ $$$${u}={e}^{{arcsinx}} \Rightarrow{u}'=\frac{{e}^{{arcsinx}} }{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${v}'=\frac{{x}}{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\Rightarrow{v}=−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${I}={xe}^{{arcsinx}} +\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{e}^{{arcsinx}} −\int{e}^{{arcsinx}} {dx}+{c}_{\mathrm{2}} \\ $$$$\mathrm{2}{I}=\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){e}^{{arcsinx}} +{c}_{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right){e}^{{arcsinx}} +{C} \\ $$