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Question Number 43699 by Meritguide1234 last updated on 14/Sep/18

Commented by Meritguide1234 last updated on 14/Sep/18

$$completeit$$

Commented by maxmathsup by imad last updated on 14/Sep/18

$$letI=\int \frac{{(1-\sqrt{1+x+x^2})}^2}{x^2\sqrt{1+x+x^2}}dxwehaveI=\int \frac{1-2\sqrt{1+x+x^2}+1+x+x^2}{x^2\sqrt{1+x+x^2}}dx$$$$=\int \frac{dx}{x^2\sqrt{1+x+x^2}}dx-2\int \frac{dx}{x^2}+\int \frac{\sqrt{1+x+x^2}}{x^2}dxbut$$$$-2\int \frac{dx}{x^2}=\frac{2}{x}+cand\int \frac{dx}{x^2\sqrt{1+x+x^2}}=\int \frac{dx}{x^2\sqrt{{(x+\frac{1}{2})}^2+\frac{3}{4}}}$$$${=}_{x+\frac{1}{2}=\frac{\sqrt{3}}{2}sh\left(t\right)}\int \frac{1}{{(\frac{\sqrt{3}}{2}sh\left(t\right)-\frac{1}{2})}^2\frac{\sqrt{3}}{2}ch\left(t\right)}\frac{\sqrt{3}}{2}ch\left(t\right)dt$$$$=\int \frac{4dt}{{(\sqrt{3}sh\left(t\right)-1)}^2}=\int \frac{4dt}{3{sh}^2\left(t\right)-2\sqrt{3}sh\left(t\right)+1}$$$$=\int \frac{4dt}{3\frac{1+ch\left(2t\right)}{2}-2\sqrt{3}sh\left(t\right)+1}=8\int \frac{dt}{3+3ch\left(2t\right)-4\sqrt{3}sh\left(t\right)+2}$$$$=8\int \frac{dt}{3+3\frac{e^{2t}+e^{-2t}}{2}-4\sqrt{3}\frac{e^t-e^{-t}}{2}+2}=16\int \frac{dt}{6+3e^{2t}+3e^{-2t}-4\sqrt{3}{e}^t+4\sqrt{3}{e}^{-t}+4}$$$${=}_{e^t=u}16\int \frac{du}{u\{10+3u^2+3u^{-2}-4\sqrt{3}u+4\sqrt{3}{u}^{-1}\}}$$$$=16\int \frac{du}{10u+3u^3+3u^{-1}-4\sqrt{3}{u}^2+4\sqrt{3}}$$$$=16\int \frac{udu}{10u^2+3u^4+3-4\sqrt{3}{u}^3+4\sqrt{3}u}letdecompose$$$$F\left(u\right)=\frac{u}{3u^4-4\sqrt{3}{u}^3+10u^2+4\sqrt{3}u+3}...becontinued...$$

Commented by MJS last updated on 14/Sep/18

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{magic}\mathrm{word}\mathrm{and}\mathrm{I}\mathrm{seem}\mathrm{to}\mathrm{remember}$$$$\mathrm{it}\mathrm{starts}\mathrm{with}‘‘{\mathrm{p}}^{″}.$$

Commented by Meritguide1234 last updated on 15/Sep/18

Answered by tanmay.chaudhury50@gmail.com last updated on 14/Sep/18

$$\int \frac{1-2\sqrt{1+x+x^2}+1+x+x^2}{x^2\sqrt{1+x+x^2}}dx$$$$\int \frac{dx}{x^2\sqrt{1+x+x^2}}-2\int \frac{dx}{x^2}+\int \frac{\sqrt{1+x+x^2}}{x^2}dx$$$$I=I_1-2I_2+I_3$$$$I_1=\int \frac{dx}{x^2\sqrt{1+x+x^2}}$$$$t=\frac{1}{x}dx$$$$dt=-\frac{dx}{x^2}$$$$\int \frac{-dt}{\sqrt{1+\frac{1}{t}+\frac{1}{t^2}}}$$$$\int \frac{-tdt}{\sqrt{t^2+t+1}}\frac{}{}$$$$\frac{-1}{2}\int \frac{2t+1-1}{\sqrt{t^2+t+1}}dt$$$$=\frac{1}{2}\int \frac{dt}{\sqrt{t^2+t+1}}-\frac{1}{2}\int \frac{d(t^2+t+1)}{\sqrt{t^2+t+1}}$$$$=\frac{1}{2}\int \frac{dt}{\sqrt{t^2+2.t.\frac{1}{2}+\frac{1}{4}+1-\frac{1}{4}}}-\frac{1}{2}\int \frac{d(t^2+t+1)}{\sqrt{t^2+t+1}}dt$$$$=\frac{1}{2}\int \frac{dt}{\sqrt{{(t+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}-\frac{1}{2}\int \frac{d(t^2+t+1)}{\sqrt{t^2+t+1}}$$$$=\frac{1}{2}ln\{(t+\frac{1}{2})+\sqrt{t^2+t+1}\}-\frac{1}{2}\times \frac{\sqrt{t^2+t+1}}{\frac{1}{2}}$$$$=\frac{1}{2}ln\{(t+\frac{1}{2})+\sqrt{t^2+t+1}\}-\sqrt{t^2+t+1}$$$$=\frac{1}{2}ln\{(\frac{1}{x}+\frac{1}{2})+\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}\}-\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}$$$$I_2=\int \frac{dx}{x^2}=\frac{x^{-1}}{-1}=\frac{-1}{x}$$$$I_3=\int \frac{\sqrt{1+x+x^2}}{x^2}$$$$=\int \frac{1+x+x^2}{x^2\sqrt{1+x+x^2}}dx$$$$=\int \frac{dx}{x^2\sqrt{1+x+x^2}}+\int \frac{dx}{x\sqrt{1+x+x^2}}+\int \frac{dx}{\sqrt{1+x+x^2}}$$$$=sameas{I}_1+\int \frac{dx}{x\sqrt{1+x+x^2}}+\int \frac{dx}{\sqrt{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}}$$$$=sameas{I}_1+ln\{(x+\frac{1}{2})+\sqrt{x^2+x+1}\}+\int \frac{dx}{x\sqrt{1+x+x^2}}$$$$=\frac{1}{2}ln\{(\frac{1}{x}+\frac{1}{2})+\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}\}-\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}+$$$$ln\{(x+\frac{1}{2}+\sqrt{x^2+x+1}\}+\int \frac{dx}{x\sqrt{1+x+x^2}}$$$$\int \frac{dx}{x\sqrt{1+x+x^2}}$$$$x=\frac{1}{t}dx=\frac{-1}{t^2}dt$$$$\int \frac{-dt}{t^2\times \frac{1}{t}\times \sqrt{1+\frac{1}{t}+\frac{1}{t^2}}}$$$$\int \frac{-dt}{\sqrt{t^2+t+1}}$$$$=-ln\{t+\frac{1}{2})+\sqrt{t^2+t+1}\}$$$$=-ln[(\frac{1}{x}+\frac{1}{2})+\left\{\sqrt{\frac{1}{x^2}+\frac{1}{x}+1}\right\}]$$$$\mathit{N}OWPLSputthevalueof{I}_1-2I_2+I_3$$$$$$$$$$$$$$

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