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Question Number 96911 by Ar Brandon last updated on 05/Jun/20

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2\sinh \left(\mathrm{x}\right)+{\tan }^{-1}\left(\mathrm{x}\right)\cdot \log ({\mathrm{x}}^4+1)}{\pi {\mathrm{e}}^{{\mathrm{x}}^2}+\sqrt[3]{{\mathrm{x}}^8+3\cosh \left(\mathrm{x}\right)}}\mathrm{dx}$$

Commented by 675480065 last updated on 05/Jun/20

$$=0$$

Commented by Ar Brandon last updated on 05/Jun/20

cool ��

Answered by Ar Brandon last updated on 05/Jun/20

$$\mathcal{I}={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2\sinh \left(\mathrm{x}\right)+{\tan }^{-1}\left(\mathrm{x}\right)\cdot \log ({\mathrm{x}}^4+1)}{\pi {\mathrm{e}}^{{\mathrm{x}}^2}+\sqrt[3]{{\mathrm{x}}^8+3\cosh \left(\mathrm{x}\right)}}\mathrm{dx}\left(1\right)$$$$\mathrm{x}=-\mathrm{u}\Rightarrow \mathrm{dx}=-\mathrm{du}$$$$\Rightarrow \mathcal{I}=\int \frac{-{\mathrm{x}}^2\sinh \left(\mathrm{x}\right)-{\tan }^{-1}\left(\mathrm{x}\right)\cdot \log ({\mathrm{x}}^4+1)}{\pi {\mathrm{e}}^{{\mathrm{x}}^2}+\sqrt{{\mathrm{x}}^8+3\cosh \left(\mathrm{x}\right)}}\mathrm{dx}\left(2\right)$$$$\left(1\right)+\left(2\right)$$$$\Rightarrow 2\mathcal{I}=0\Rightarrow \mathcal{I}=0$$

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