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Question Number 128109 by bemath last updated on 04/Jan/21

$$\mathrm{\varnothing }=\int (\mathrm{x}-2)\sqrt{\frac{\mathrm{x}+1}{\mathrm{x}-1}}\mathrm{dx}$$

Answered by liberty last updated on 04/Jan/21

$$\varphi =\int \frac{(\mathrm{x}-2)\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}-1}\mathrm{dx}$$$$\mathrm{let}\mathrm{x}=\sec \mathrm{t}$$$$\mathrm{\varnothing }=\int \frac{(\sec \mathrm{t}-2).\tan \mathrm{t}}{\sec \mathrm{t}-1}.\left(\sec \mathrm{t}\tan \mathrm{t}\right)\mathrm{dt}$$$$\varphi =\int \frac{\left(\frac{1-2\cos \mathrm{t}}{\cos \mathrm{t}}\right).\frac{\sin {}^2\mathrm{t}}{\cos {}^3\mathrm{t}}}{\frac{1-\cos \mathrm{t}}{\cos \mathrm{t}}}\mathrm{dt}$$$$\varphi =\int \frac{(1-2\cos \mathrm{t})\sin {}^2\mathrm{t}}{\cos {}^3\mathrm{t}(1-\cos \mathrm{t})}\mathrm{dt}$$$$\varphi =\int \frac{(1-2\cos \mathrm{t})(1-\cos {}^2\mathrm{t})}{\cos {}^3\mathrm{t}(1-\cos \mathrm{t})}\mathrm{dt}$$$$\varphi =\int \frac{(1-2\cos \mathrm{t})(1+\cos \mathrm{t})}{\cos {}^3\mathrm{t}}\mathrm{dt}$$$$\varphi =\int \frac{1-\cos \mathrm{t}-2\cos {}^2\mathrm{t}}{\cos {}^3\mathrm{t}}\mathrm{dt}$$$$\mathrm{\varnothing }=\int \sec {}^3\mathrm{t}\mathrm{dt}-\int \sec {}^2\mathrm{t}\mathrm{dt}-2\int \sec \mathrm{t}\mathrm{dt}$$$$\mathrm{\varnothing }=\frac{1}{2}\sec \mathrm{t}\tan \mathrm{t}+\frac{1}{2}\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid -\tan \mathrm{t}-2\ln \mid \sec \mathrm{t}+\tan \mathrm{t}\mid +\mathrm{C}$$$$\varphi =\frac{1}{2}\mathrm{x}\sqrt{{\mathrm{x}}^2-1}-\sqrt{{\mathrm{x}}^2-1}-\frac{3}{2}\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2-1}\mid +\mathrm{C}$$$$\varphi =\frac{(\mathrm{x}-1)\sqrt{{\mathrm{x}}^2-1}}{2}-\frac{3}{2}\ln \mid \mathrm{x}+\sqrt{{\mathrm{x}}^2-1}\mid +\mathrm{C}$$

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