Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 41302 by math khazana by abdo last updated on 05/Aug/18

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

Commented by maxmathsup by imad last updated on 05/Aug/18

$$letI={\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2\sqrt{x^2+x+1}}$$$$I={\int }_1^{+\mathrm{\infty }}\frac{dx}{x^2\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}changementx+\frac{1}{2}=\frac{\sqrt{3}}{2}tan\theta give$$$$I={\int }_{arctan\left(\frac{1}{\sqrt{3}}\right)}^{\frac{\pi }{2}}\frac{1}{{(\frac{\sqrt{3}}{2}tan\theta -\frac{1}{2})}^2\frac{\sqrt{3}}{2}}{cos}^2\theta \frac{\sqrt{3}}{2}(1+{tan}^2\theta )d\theta$$$$={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{4d\theta }{{(\sqrt{3}tan\theta -1)}^2}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{4}{3{tan}^2\theta -2\sqrt{3}tan\theta +1}d\theta$$$${=}_{tan\theta =x}{\int }_{\frac{1}{\sqrt{3}}}^{+\mathrm{\infty }}\frac{4}{(3x^2-2\sqrt{3x}+1)}\frac{dx}{1+x^2}letdecompose$$$$F\left(x\right)=\frac{4}{(x^2+1)(3x^2-2\sqrt{3}x+1)}=\frac{4}{{(\sqrt{3}x-1)}^2(x^2+1)}$$$$F\left(x\right)=\frac{a}{\sqrt{3}x-1}+\frac{b}{{(\sqrt{3}x-1)}^2}+\frac{bx+c}{x^2+1}$$$$b={lim}_{x\to \frac{1}{\sqrt{3}}}{(\sqrt{3}x-1)}^{2}F\left(x\right)=\frac{4}{\frac{4}{3}}=3$$$${lim}_{x\to +\mathrm{\infty }}xF\left(x\right)=0=\frac{a}{\sqrt{3}}+b\Rightarrow b=-\frac{a}{\sqrt{3}}\Rightarrow$$$$F\left(x\right)=\frac{a}{\sqrt{3}x-1}+\frac{3}{{(\sqrt{3}x-1)}^2}+\frac{-\frac{a}{\sqrt{3}}x+c}{x^2+1}$$$$F\left(0\right)=-a+3+c\Rightarrow c=a-3\Rightarrow$$$$F\left(x\right)=\frac{a}{\sqrt{3}x-1}+\frac{3}{{(\sqrt{3}x-1)}^2}+\frac{-\frac{a}{\sqrt{3}}x+a-3}{x^2+1}$$$$F\left(\sqrt{3}\right)=\frac{4}{4\left(4\right)}=\frac{1}{4}=\frac{a}{2}+\frac{3}{4}+\frac{-3}{4}\Rightarrow \frac{a}{2}=\frac{1}{4}\Rightarrow a=\frac{1}{2}\Rightarrow$$$$F\left(x\right)=\frac{1}{2(\sqrt{3}x-1)}+\frac{3}{{(\sqrt{3}x-1)}^2}+\frac{-\frac{1}{2\sqrt{3}}x-\frac{5}{2}}{x^2+1}\Rightarrow$$$$\int F\left(x\right)dx=\frac{1}{2\sqrt{3}}\int \frac{dx}{x-\frac{1}{\sqrt{3}}}+\int \frac{dx}{{(x-\frac{1}{\sqrt{3}})}^2}-\frac{1}{2\sqrt{3}}\int \frac{x+5\sqrt{3}}{x^2+1}dx$$$$=\frac{1}{2\sqrt{3}}ln\mid x-\frac{1}{\sqrt{3}}\mid -\frac{1}{x-\frac{1}{\sqrt{3}}}-\frac{1}{4\sqrt{3}}\int \frac{2x}{x^2+1}-\frac{5}{2}\int \frac{dx}{x^2+1}$$$$=\frac{1}{2\sqrt{3}}ln\mid x-\frac{1}{\sqrt{3}}\mid -\frac{1}{x-\frac{1}{\sqrt{3}}}-\frac{1}{4\sqrt{3}}ln(x^2+1)-\frac{5}{2}arctanx+c\Rightarrow$$$${\int }_{\frac{1}{\sqrt{3}}}^{+\mathrm{\infty }}F\left(x\right)dx=....\left(putthelmits\right)$$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Aug/18

$${\int }_1^{\mathrm{\infty }}\frac{\frac{1}{x^2}}{\sqrt{x^2(1+\frac{1}{x}+\frac{1}{x^2})}}dx$$$$t=\frac{1}{x}dt=-\frac{1}{x^2}dx$$$${\int }_1^0\frac{-dt}{\sqrt{\frac{1}{t^2}(1+t+t^2)}}$$$${\int }_0^1\frac{tdt}{\sqrt{t^2+t+1}}$$$$=\frac{1}{2}{\int }_0^1\frac{2t+1-1}{\sqrt{t^2+t+1}}dt$$$$=\frac{1}{2}{\int }_0^1\frac{2t+1}{\sqrt{t^2+t+1}}-\frac{1}{2}{\int }_0^1\frac{dt}{\sqrt{t^2+2.t.\frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}}}$$$$=\frac{1}{2}{\int }_0^1\frac{d(t^2+t+1)}{\sqrt{t^2+t+1}}-\frac{1}{2}{\int }_0^1\frac{dt}{\sqrt{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$$$$=\frac{1}{2}\mid \frac{\sqrt{t^2+t+1}}{\frac{1}{2}}{\mid }_0^1-\frac{1}{2}\mid ln\{(t+\frac{1}{2})+\sqrt{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\}{\mid }_0^1$$$$=(\sqrt{3}-1)-\frac{1}{2}\{ln(\frac{3}{2}+\sqrt{\frac{3}{4}+\frac{3}{4}})-ln(\frac{1}{2}+1)\}$$$$=(\sqrt{3}-1)-\frac{1}{2}\{ln\left(\frac{3+\sqrt{6}}{2}\right)-ln\left(\frac{3}{2}\right)\}$$$$=(\sqrt{3}-1)-\frac{1}{2}\left\{ln\left(\frac{3+\sqrt{6}}{3}\right)\right\}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com