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Question Number 97537 by  M±th+et+s last updated on 08/Jun/20

$${\int }_0^1\frac{-\sqrt{1-x^2}}{({yx}^3+x^2-yx-1)}dx$$

Answered by smridha last updated on 08/Jun/20

$${\int }_0^1\frac{dx}{(\mathit{y}\mathit{x}+1)\sqrt{1-{\mathit{x}}^2}}\mathit{l}\mathit{e}\mathit{t}\mathit{x}=\mathit{s}\mathit{i}\mathit{n}\mathit{A}$$$$={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{\mathit{d}\mathit{A}}{\mathit{y}\mathit{s}\mathit{i}\mathit{n}\mathit{A}+1}=2{\int }_0^{\frac{\pi }{2}}\frac{\mathit{d}(\mathit{t}\mathit{a}\mathit{n}\frac{\mathit{A}}{2}+\mathit{y})}{{(\mathit{t}\mathit{a}\mathit{n}\frac{\mathit{A}}{2}+1)}^2+{\left(\sqrt{1-{\mathit{y}}^2}\right)}^2}$$$$=2.\frac{1}{\sqrt{1-{\mathit{y}}^2}}{\left[{\tan }^{-1}\left(\frac{\mathit{t}\mathit{a}\mathit{n}\frac{\mathit{A}}{2}+y}{\sqrt{1-{\mathit{y}}^2}}\right)\right]}_0^{\frac{\mathit{\pi }}{2}}$$$$=\frac{2}{\sqrt{1-{\mathit{y}}^2}}[{\tan }^{-1}\left(\frac{1+\mathit{y}}{\sqrt{1-{\mathit{y}}^2}}\right)-{\tan }^{-1}\left(\frac{\mathit{y}}{\sqrt{1-{\mathit{y}}^2}}\right)$$$$=\frac{1}{\sqrt{1-{\mathit{y}}^2}}{\tan }^{-1}\left[\frac{(1+\mathit{y})\sqrt{1-{\mathit{y}}^2}}{\mathit{y}}\right]$$

Commented by  M±th+et+s last updated on 08/Jun/20

$$thankyousir$$

Commented by smridha last updated on 08/Jun/20

welcome

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