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Question Number 205670 by hardmath last updated on 26/Mar/24

Answered by Berbere last updated on 27/Mar/24

$$\mathrm{\Omega }={\int }_1^{\sqrt{3}}{\left(\frac{{\tan }^{-1}\left(x\right)}{x-{\tan }^{-1}\left(x\right)}\right)}^{2}dx$$$${\int }_1^{\sqrt{3}}(1+\frac{x^2}{{(x-{\tan }^{-1}\left(x\right))}^2}-\frac{2x}{(x-{\tan }^{-1}\left(x\right))})dx$$$$=(\sqrt{3}-1)+{\int }_1^{\sqrt{3}}\frac{-x^2+2x{\tan }^{-1}\left(x\right)}{{(x-{\tan }^{-1}\left(x\right))}^2}dx..E$$$$\frac{1}{(x-{\tan }^{-1}\left(x\right))}=u;u^{\prime }=\frac{-\frac{x^2}{1+x^2}}{{(x-{\tan }^{-1}\left(x\right))}^2}$$$$E=(\sqrt{3}-1)+{\int }_1^{\sqrt{3}}\frac{\frac{x^2}{(1+x^2)}.(1+x^2)-{(1+x^2)}^{\prime }.(-{\tan }^{-1}\left(x\right)+x)}{{(x-{\tan }^{-1}\left(x\right))}^2}dx$$$$=(\sqrt{3}-1)+{\int }_1^{\sqrt{3}}(\frac{-{({\tan }^{-1}\left(x\right)-x)}^{\prime }.(1+x^2)+{(1+x^2)}^{\prime }({\tan }^{-1}\left(x\right)-x)}{(x-{\tan }^{-1}\left(x\right))}dx$$$$\sqrt{3}-1+{\left[\frac{x^2+1}{{\tan }^{-1}\left(x\right)-x}\right]}_1^{\sqrt{3}}=\sqrt{3}-1+\frac{12}{\pi -3\sqrt{3}}-\frac{8}{\pi -4}$$$$$$$$$$$$$$$$$$

Commented by hardmath last updated on 27/Mar/24

$$\mathrm{Thank}\mathrm{you}\mathrm{dear}\mathrm{professor}\mathrm{cool}$$

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