Question Number 202988 by sonukgindia last updated on 06/Jan/24
Commented by Frix last updated on 06/Jan/24
$$\mathrm{By}\:\mathrm{parts}. \\ $$
Answered by MM42 last updated on 07/Jan/24
$${lnx}={u}\Rightarrow\frac{{dx}}{{x}}={du} \\ $$$${sin}^{−\mathrm{1}} {xdx}={dv}\Rightarrow{xsin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }={v} \\ $$$$\Rightarrow{I}={xsin}^{−\mathrm{1}} {xlnx}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }{lnx} \\ $$$$−\int\:{sin}^{−\mathrm{1}} {xdx}−\int\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}{dx} \\ $$$$\bigstar\:\int\:\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}{dx}\overset{{x}={sinu}} {\rightarrow}=\int\:\frac{{cos}^{\mathrm{2}} {u}}{{sinu}}\:{du} \\ $$$$=\int\:\left(\frac{\mathrm{1}}{{sinu}}−{sinu}\right){du}={ln}\left({tan}\frac{{u}}{\mathrm{2}}\right)+{cosu} \\ $$$$=\frac{\mathrm{1}−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{I}=\left({lnx}−\mathrm{1}\right)\left({xsin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:\checkmark \\ $$$$ \\ $$
Commented by MathematicalUser2357 last updated on 11/Jan/24
$$\left(\mathrm{ln}\:{x}−\mathrm{1}\right)\left({x}\mathrm{sin}^{−\mathrm{1}} {x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)−\frac{\mathrm{1}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }+{C} \\ $$