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Question Number 94119 by i jagooll last updated on 17/May/20

$$\int {\cot }^{-1}\left(\sqrt{\mathrm{x}}\right)\mathrm{dx}$$

Commented by i jagooll last updated on 17/May/20

$$u={\cot }^{-1}\left(\sqrt{x}\right)\Rightarrow du=\frac{dx}{2\sqrt{x}(1-x)}$$$$v=x$$$$\int {\cot }^{-1}\left(\sqrt{x}\right)dx=x{\cot }^{-1}\left(\sqrt{x}\right)$$$$-\frac{1}{2}\int \frac{x}{\sqrt{x}(1-x)}dx$$$$=x{\cot }^{-1}\left(\sqrt{x}\right)+\frac{1}{2}\int \frac{\sqrt{x}}{x-1}dx$$$$=x{\cot }^{-1}\left(\sqrt{x}\right)+\frac{1}{2}\int \frac{\sqrt{x}}{{\left(\sqrt{x}\right)}^2-1}dx$$$$\mathrm{let}\sqrt{x}=\tan j\iff dx=2\tan {}^{2}j\sec {}^{2}jdj$$$$\frac{1}{2}\int \frac{2\tan {}^{3}j\sec {}^{2}jdj}{\sec {}^{2}j}=$$$$\int \tan j(\sec {}^{2}j-1)dj=$$$$\frac{1}{2}\tan j+\ln \left(\cos j\right)+c$$$$\frac{1}{2}\sqrt{x}+\ln \left(\frac{1}{x+1}\right)$$$$\therefore x{\cot }^{-1}\left(\sqrt{x}\right)+\frac{1}{2}\sqrt{x}+\ln \left(\frac{1}{x+1}\right)+c$$

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