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Question Number 51324 by Abdo msup. last updated on 25/Dec/18

$$calculate{\int }_0^{+\mathrm{\infty }}\frac{dx}{1+x^2+x^4}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Dec/18

$$\int \frac{dx}{x^4+x^2+1}$$$$\frac{1}{2}\int \frac{\frac{2}{x^2}}{x^2+1+\frac{1}{x^2}}$$$$\frac{1}{2}\int \frac{(1+\frac{1}{x^2})-(1-\frac{1}{x^2})}{(x^2+\frac{1}{x^2}+1)}dx$$$$\frac{1}{2}\int \frac{d(x-\frac{1}{x})}{{(x-\frac{1}{x})}^2+3}-\frac{1}{2}\int \frac{d(x+\frac{1}{x})}{{(x+\frac{1}{x})}^2-1}$$$$\frac{1}{2}\times \frac{1}{\sqrt{3}}{tan}^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{3}}\right)-\frac{1}{2}\times \frac{1}{2}ln\left(\frac{x+\frac{1}{x}-1}{x+\frac{1}{x}+1}\right)+c_1$$$$=\frac{1}{2\sqrt{3}}\times {tan}^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right)-\frac{1}{4}ln\left(\frac{x^2-x+1}{x^2+x+1}\right)+c_1$$$$requiredanswer$$$$=\mid \frac{1}{2\sqrt{3}}{tan}^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right)-\frac{1}{4}ln\left(\frac{1-\frac{1}{x}+\frac{1}{x^2}}{1+\frac{1}{x}+\frac{1}{x^2}}\right){\mid }_0^{\mathrm{\infty }}$$$$=\mid \frac{1}{2\sqrt{3}}{tan}^{-1}\left(\frac{1-\frac{1}{x^2}}{\frac{\sqrt{3}}{x}}\right)-do{\mid }_0^{\mathrm{\infty }}$$$$=\frac{1}{2\sqrt{3}}[{tan}^{-1}\left(\frac{1-0}{0}\right)-{tan}^{-1}\left(\frac{0^2-1}{0\times \sqrt{3}}\right)]-\frac{1}{4}[ln\left(\frac{1-0+0}{1+0+0}\right)-ln\left(\frac{0^2-0+1}{0^2+0+1}\right)]$$$$=\frac{1}{2\sqrt{3}}[tan\left(\mathrm{\infty }\right)-{tan}^{-1}(-\mathrm{\infty })]-do$$$$=\frac{1}{2\sqrt{3}}[\frac{\pi }{2}-\left(\frac{-\pi }{2}\right)]$$$$=\frac{\pi }{2\sqrt{3}}$$$$$$

Commented by peter frank last updated on 26/Dec/18

$$niceworksir$$

Answered by Smail last updated on 26/Dec/18

$$\frac{1}{(x^2+x+1)(x^2-x+1)}=\frac{ax+b}{x^2+x+1}+\frac{cx+d}{x^2-x+1}$$$$a=-c,d=1-b$$$$b=1-a$$$$a=b=-c=d=\frac{1}{2}$$$$A={\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2+x^4}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x+1}{x^2+x+1}dx-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{x-1}{x^2-x+1}dx$$$$=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{2x+1+1}{x^2+x+1}dx-\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{2x-1-1}{x^2-x+1}dx$$$$=\frac{1}{4}{[ln\mid \frac{x^2+x+1}{x^2-x+1}\mid ]}_0^{\mathrm{\infty }}+\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x+\frac{1}{2})}^2+\frac{3}{4}}+\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x-\frac{1}{2})}^2+\frac{3}{4}}$$$$=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{dx}{{\left(\frac{2x+1}{\sqrt{3}}\right)}^2+1}+\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{dx}{{\left(\frac{2x-1}{\sqrt{3}}\right)}^2+1}$$$$t=\frac{2x\underset{-}{+}1}{\sqrt{3}}\Rightarrow dx=\frac{\sqrt{3}}{2}dt$$$$A=\frac{1}{2\sqrt{3}}{\int }_{1/\sqrt{3}}^{\mathrm{\infty }}\frac{dt}{t^2+1}+\frac{1}{2\sqrt{3}}{\int }_{-1/\sqrt{3}}^{\mathrm{\infty }}\frac{dt}{t^2+1}$$$$=\frac{1}{2\sqrt{3}}{\left[{tan}^{-1}\left(t\right)\right]}_{1/\sqrt{3}}^{\mathrm{\infty }}+\frac{1}{2\sqrt{3}}{\left[{tan}^{-1}\left(t\right)\right]}_{-1/\sqrt{3}}^{\mathrm{\infty }}$$$$=\frac{1}{2\sqrt{3}}(\frac{\pi }{2}-{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right))+\frac{1}{2\sqrt{3}}(\frac{\pi }{2}+{tan}^{-1}\left(\frac{\sqrt{3}}{3}\right))$$$$=\frac{\pi }{2\sqrt{3}}$$

Commented by peter frank last updated on 26/Dec/18

$$niceworksir$$

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