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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

$$lnx=u\Rightarrow \frac{dx}{x}=du$$$${sin}^{-1}xdx=dv\Rightarrow {xsin}^{-1}x+\sqrt{1-x^2}=v$$$$\Rightarrow I={xsin}^{-1}xlnx+\sqrt{1-x^2}lnx$$$$-\int {sin}^{-1}xdx-\int \frac{\sqrt{1-x^2}}{x}dx$$$$★\int \frac{\sqrt{1-x^2}}{x}dx\stackrel{x=sinu}{\to }=\int \frac{{cos}^{2}u}{sinu}du$$$$=\int (\frac{1}{sinu}-sinu)du=ln\left(tan\frac{u}{2}\right)+cosu$$$$=\frac{1-\sqrt{1-x^2}}{x^2}+\sqrt{1-x^2}$$$$\Rightarrow I=(lnx-1)({xsin}^{-1}x+\sqrt{1-x^2})-\frac{1+\sqrt{1-x^2}}{x^2}-\sqrt{1-x^2}✓$$$$$$

Commented by MathematicalUser2357 last updated on 11/Jan/24

$$(\ln x-1)(x{\sin }^{-1}x+\sqrt{1-x^2})-\frac{1+\sqrt{1-x^2}}{x^2}-\sqrt{1-x^2}+C$$

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