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Question Number 31967 by abdo imad last updated on 17/Mar/18

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{arctanx}{x^2+x+1}dx.$$

Commented by abdo imad last updated on 19/Mar/18

$$letputI={\int }_0^{\mathrm{\infty }}\frac{arctanx}{x^2+x+1}dx.ch.x=\frac{1}{t}give$$$$I={\int }_0^{\mathrm{\infty }}\frac{\frac{\pi }{2}-arctant}{\frac{1}{t^2}+\frac{1}{t}+1}\frac{dt}{t^2}={\int }_0^{\mathrm{\infty }}\frac{\frac{\pi }{2}-arctant}{1+t+t^2}dt$$$$=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+t+1}-I\Rightarrow 2I=\frac{\pi }{2}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+t+1}\Rightarrow$$$$I=\frac{\pi }{4}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+t+1}.but$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+t+1}={\int }_0^{\mathrm{\infty }}\frac{dt}{{(t+\frac{1}{2})}^2+\frac{3}{4}}(.ch.t+\frac{1}{2}=\frac{\sqrt{3}}{2}u)$$$$={\int }_{\frac{1}{\sqrt{3}}}^{+\mathrm{\infty }}\frac{1}{\frac{3}{4}(t^2+1)}\frac{\sqrt{3}}{2}du=\frac{4}{3}\frac{\sqrt{3}}{2}{\int }_{\frac{1}{\sqrt{3}}}^{+\mathrm{\infty }}\frac{du}{1+u^2}$$$$=\frac{2\sqrt{3}}{3}{\left[arctan\left(u\right)\right]}_{\frac{1}{\sqrt{3}}}^{\mathrm{\infty }}=\frac{2\sqrt{3}}{3}(\frac{\pi }{2}-arctan\left(\frac{1}{\sqrt{3}}\right))$$$$=\frac{2\sqrt{3}}{3}arctan\left(\sqrt{3}\right)=\frac{2\sqrt{3}}{3}\frac{\pi }{3}=\frac{2\pi \sqrt{3}}{9}\Rightarrow$$$$I=\frac{\pi }{4}\frac{2\pi \sqrt{3}}{9}=\frac{{\pi }^2\sqrt{3}}{18}.$$

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