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Question Number 33202 by prof Abdo imad last updated on 12/Apr/18

$$findthevalueof{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(1+t+t^2)}^2}.$$

Commented by prof Abdo imad last updated on 13/Apr/18

$$letputI={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{(1+t+t^2)}^2}wehave$$$$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{dt}{{({(t+\frac{1}{2})}^2+\frac{3}{4})}^2}.changement$$$$t+\frac{1}{2}=\frac{\sqrt{3}}{2}tan\theta give$$$$I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{{(\frac{3}{4}{tan}^2\theta +\frac{3}{4})}^2}\frac{\sqrt{3}}{2}(1+{tan}^2\theta )d\theta$$$$=\frac{\sqrt{3}}{2}.\frac{16}{9}{\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{d\theta }{1+{tan}^2\theta }=\frac{16\sqrt{3}}{9}{\int }_0^{\frac{\pi }{2}}{cos}^2\theta d\theta$$$$=\frac{16\sqrt{3}}{9}{\int }_0^{\frac{\pi }{2}}\frac{1+cos\left(2\theta \right)}{2}d\theta$$$$=\frac{8\sqrt{3}}{9}{\int }_0^{\frac{\pi }{2}}(1+cos\left(2\theta \right))d\theta$$$$=\frac{4\sqrt{3}\pi }{9}+\frac{4\sqrt{3}}{9}{\left[sin\left(2\theta \right)\right]}_0^{\frac{\pi }{2}}=\frac{4\pi \sqrt{3}}{9}+0\Rightarrow$$$$I=\frac{4\pi \sqrt{3}}{9}.$$

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