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Question Number 129377 by pipin last updated on 15/Jan/21

$$\int \frac{\sqrt{\mathbf{x}}}{\sqrt{\mathbf{x}-1}}\mathrm{dx}=...$$

Answered by Ar Brandon last updated on 15/Jan/21

$$\mathcal{I}=\int \frac{\sqrt{\mathrm{x}}}{\sqrt{\mathrm{x}-1}}\mathrm{dx},\mathrm{x}={\mathrm{t}}^2\Rightarrow \mathrm{dx}=2\mathrm{tdt}$$$$=2\int \frac{{\mathrm{t}}^2}{\sqrt{{\mathrm{t}}^2+1}}\mathrm{dt}=2\int \{\sqrt{{\mathrm{t}}^2+1}-\frac{1}{\sqrt{{\mathrm{t}}^2+1}}\}\mathrm{dt}$$

Answered by liberty last updated on 15/Jan/21

$$\mathrm{let}\sqrt{\mathrm{x}-1}=\tan \mathrm{t};\mathrm{x}=\sec {}^2\mathrm{t};\mathrm{dx}=2\sec {}^2\mathrm{t}\tan \mathrm{t}\mathrm{dt}$$$$\mathrm{D}=\int \frac{\sec \mathrm{t}}{\tan \mathrm{t}}\left(2\sec {}^2\mathrm{t}\tan \mathrm{t}\right)\mathrm{dt}$$$$\mathrm{D}=2\int \sec {}^3\mathrm{t}\mathrm{dt}=2(\frac{1}{2}\sec \mathrm{t}\tan \mathrm{t}+\frac{1}{2}\ln \mid \tan (\frac{\mathrm{t}}{2}+\frac{\pi }{4})\mid )+\mathrm{C}$$$$\mathrm{D}=\sqrt{\mathrm{x}}\sqrt{\mathrm{x}-1}+\ln \mid \tan (\frac{{\tan }^{-1}\left(\sqrt{\mathrm{x}-1}\right)}{2}+\frac{\pi }{4})\mid +\mathrm{C}$$$$\mathrm{D}=\sqrt{{\mathrm{x}}^2-1}+\ln \mid \tan (\frac{{\tan }^{-1}\left(\sqrt{\mathrm{x}-1}\right)}{2}+\frac{\pi }{4})\mid +\mathrm{C}$$

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