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Question Number 193734 by Mastermind last updated on 18/Jun/23

$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{{\mathrm{x}}^2{\left({\tan }^{-1}\left(\mathrm{x}\right)\right)}^2}{{\mathrm{x}}^2-\mathrm{x}+2}\mathrm{dx}$$$$$$$$\mathrm{Help}!$$

Answered by witcher3 last updated on 18/Jun/23

$$\mathrm{I}\mathrm{didnt}\mathrm{Converge}$$$$\mathrm{x}>1$$$$\frac{{\mathrm{x}}^2{\left({\tan }^{-}\left(\mathrm{x}\right)\right)}^2}{{\mathrm{x}}^2-\mathrm{x}+2}⩾\frac{{\pi }^2}{16}.\frac{{\mathrm{x}}^2}{{\mathrm{x}}^2-\mathrm{x}+2}⩾\frac{{\pi }^2}{16}.\frac{2\mathrm{x}-1}{{\mathrm{x}}^2-\mathrm{x}+2},{\mathrm{x}}^2⩾2\mathrm{x}-1$$$$\mathrm{I}⩾\frac{{\pi }^2}{16}{\left[\ln ({\mathrm{x}}^2-\mathrm{x}+2)\right]}_0^{\mathrm{a}},\mathrm{\forall }\mathrm{a}⩾1$$$$\Rightarrow \mathrm{I}⩾\underset{\mathrm{a}\to \mathrm{\infty }}{\lim }\frac{{\pi }^2}{16}\ln \left[\frac{{\mathrm{a}}^2-\mathrm{a}+2}{2}\right]=\mathrm{\infty }$$$$$$

Commented by Mastermind last updated on 20/Jun/23

$$\mathrm{Thank}\mathrm{you}$$

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