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Question Number 97059 by mhmd last updated on 06/Jun/20

Commented by PRITHWISH SEN 2 last updated on 06/Jun/20

$$\mathrm{let}$$$$\frac{{\mathrm{x}}^2+1}{{\sin }^{-1}\mathrm{x}-1}=\mathrm{v}\left(\mathrm{x}\right)$$$$\mathrm{then}{\mathrm{v}}^{{}^{\prime }}\left(\mathrm{x}\right)=\frac{{2\mathrm{xsin}}^{-1}\mathrm{x}.\sqrt{1-{\mathrm{x}}^2}-2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}-{\mathrm{x}}^2-1}{\sqrt{1-{\mathrm{x}}^2}{({\sin }^{-1}\mathrm{x}-1)}^2}$$$$\int \{{\mathrm{e}}^{\mathrm{x}}\mathrm{v}\left(\mathrm{x}\right)+{\mathrm{e}}^{\mathrm{x}}{\mathrm{v}}^{\prime }\left(\mathrm{x}\right)\}\mathrm{dx}$$$$\mathrm{using}\mathrm{by}\mathrm{parts}$$$${\mathrm{e}}^{\mathrm{x}}\mathrm{v}\left(\mathrm{x}\right)-\int {\mathrm{e}}^{\mathrm{x}}{\mathrm{v}}^{\prime }\left(\mathrm{x}\right)+\int {\mathrm{e}}^{\mathrm{x}}{\mathrm{v}}^{{}^{\prime }}\left(\mathrm{x}\right)+\mathrm{C}$$$$={\mathrm{e}}^{\mathrm{x}}\mathrm{v}\left(\mathrm{x}\right)+\mathrm{C}={\mathbf{e}}^{\mathbf{x}}.\frac{{\mathbf{x}}^2+1}{{\sin }^{-1}\mathbf{x}-1}+\mathrm{C}\mathbf{please}\mathbf{check}$$

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