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Question Number 64866 by mathmax by abdo last updated on 22/Jul/19

$$find{\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2\sqrt{1+x+x^2}}$$

Commented by mathmax by abdo last updated on 23/Jul/19

$$letA={\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2\sqrt{x^2+x+1}}wehave{x}^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}$$$$changementx+\frac{1}{2}=\frac{\sqrt{3}}{2}sh\left(t\right)givesh\left(t\right)=\frac{2x+1}{\sqrt{3}}$$$$A={\int }_{argsh\left(\sqrt{3}\right)}^{+\mathrm{\infty }}\frac{1}{{(\frac{\sqrt{3}}{2}sht-\frac{1}{2})}^2\frac{\sqrt{3}}{2}ch\left(t\right)}\frac{\sqrt{3}}{2}ch\left(t\right)dt$$$$=4{\int }_{ln(\sqrt{3}+\sqrt{4})}^{+\mathrm{\infty }}\frac{dt}{{(\sqrt{3}sh\left(t\right)-1)}^2}=4{\int }_{ln(2+\sqrt{3})}^{+\mathrm{\infty }}\frac{dt}{3{sh}^{2}t-2\sqrt{3}sht+1}$$$$=4{\int }_{ln(2+\sqrt{3})}^{+\mathrm{\infty }}\frac{dt}{3\frac{ch\left(2t\right)-1}{2}-2\sqrt{3}sh\left(t\right)+1}$$$$=8{\int }_{ln(2+\sqrt{3})}^{+\mathrm{\infty }}\frac{dt}{3ch\left(2t\right)-3-4\sqrt{3}sh\left(t\right)+2}$$$$=8{\int }_{ln(2+\sqrt{3})}^{+\mathrm{\infty }}\frac{dt}{3ch\left(2t\right)-4\sqrt{3}sh\left(t\right)-1}$$$$=8{\int }_{ln(2+\sqrt{3})}^{+\mathrm{\infty }}\frac{dt}{3\frac{e^{2t}+e^{-2t}}{2}-4\sqrt{3}\frac{e^t-e^{-t}}{2}-1}$$$$=16{\int }_{ln(2+\sqrt{3})}^{+\mathrm{\infty }}\frac{dt}{3e^{2t}+3e^{-2t}-4\sqrt{3}{e}^t+4\sqrt{3}{e}^{-t}-2}$$$${=}_{e^t=u}16{\int }_{2+\sqrt{3}}^{+\mathrm{\infty }}\frac{du}{u(3u^2+3u^{-2}-4\sqrt{3}u+4\sqrt{3}{u}^{-1}-2)}$$$$=16{\int }_{2+\sqrt{3}}^{+\mathrm{\infty }}\frac{du}{3u^3+3u^{-1}-4\sqrt{3}{u}^2+4\sqrt{3}-2u}$$$$=16{\int }_{2+\sqrt{3}}^{+\mathrm{\infty }}\frac{udu}{3u^4+3-4\sqrt{3}{u}^3+4\sqrt{3}u-2u^2}letdecompose$$$$F\left(u\right)=\frac{u}{3u^4-4\sqrt{3}{u}^3-2u^2+4\sqrt{3}u+3}....becontinued....$$$$$$

Commented by ~ À ® @ 237 ~ last updated on 25/Jul/19

$$letuschangeu=\frac{1}{x}du=-\frac{dx}{x^2}$$$$thenwehaveI={\int }_0^1\frac{udu}{\sqrt{1+u+u^2}}=\frac{1}{2}[{\int }_0^1\frac{2u+1}{\sqrt{1+u+u^2}}du-{\int }_0^1\frac{du}{\sqrt{1+u+u^2}}]$$$$forthe{2}^{nd}integralJweknowsthat1+u+u^2=\frac{3}{4}[{\left(\frac{2u+1}{\sqrt{3}}\right)}^2+1]$$$$letuschangev=arctan\left(\frac{2u+1}{\sqrt{3}}\right)dv=\frac{\frac{2}{\sqrt{3}}du}{{\left(\frac{2u+1}{\sqrt{3}}\right)}^2+1}$$$$so$$$$J={\int }_0^1\frac{du}{\sqrt{1+u+u^2}}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\sqrt{{tan}^{2}v+1}dv={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dv}{cosv}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{cosvdv}{(1-sinv)(1+sinv)}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{1}{2}.[\frac{cosv}{1-sinv}+\frac{cosv}{1+sinv}]dv$$$$={[\frac{1}{2}ln\mid \frac{1+sinv}{1-sinv}\mid ]}_{\frac{\pi }{6}}^{\frac{\pi }{3}}=ln\left(\frac{4+2\sqrt{3}}{3}\right)$$$$Now$$$$I=\frac{1}{2}.({\left[2\sqrt{1+u+u^2}\right]}_0^1-J)=\frac{1}{2}.[(2\sqrt{3}-2)-ln\left(\frac{4+2\sqrt{3}}{3}\right)]$$$$$$

Commented by mathmax by abdo last updated on 28/Jul/19

$$thankyousir.$$

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