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Question Number 176485 by Shrinava last updated on 20/Sep/22

Commented by mr W last updated on 20/Sep/22

$$x=4isthesolution.$$

Answered by a.lgnaoui last updated on 20/Sep/22

$$\int \frac{t^2+1}{t^3+1}dt=\int \frac{t^2}{t^3+1}+\int \frac{1}{t^3+1}dt$$$$=\frac{1}{2}\int \frac{2t^2}{1+t^3}dt+\int \frac{1}{(t+1)(t^2-t+1)}dt$$$$\int \frac{1}{t+1)(t^2-t+1)}dt=\frac{1}{3}\int (\frac{1}{t+1}+\frac{2-t}{t^2-t+1})dt$$$$\int \frac{2-t}{t^2-t+1}dt=\int \frac{-(t-\frac{1}{2})+\frac{3}{2}}{{(t-\frac{1}{2})}^2+\frac{3}{4}}dt$$$$=-\frac{1}{2}\int \frac{2t-1}{t^2-t+1}+\frac{3}{2}\times \frac{4}{3}\int \frac{dt}{{\left[\frac{2}{\sqrt{3}}(t-\frac{1}{2})\right]}^2+1}$$$$=-\frac{1}{2}\log ({\mathrm{t}}^2-\mathrm{t}+1)+2\mathrm{Arctan}\left[\frac{\sqrt{3}}{3}(2t-1)\right]$$$$finalement$$$$I=\frac{1}{2}\log (1+{\mathrm{t}}^3)+\frac{1}{3}\log (1+\mathrm{t})-\frac{1}{2}\log ({\mathrm{t}}^2-\mathrm{t}+1)+\frac{1}{3}2\mathrm{Arctan}[\frac{\sqrt{3}}{3}(2\mathrm{t}-1)$$$${\int }_4^x\frac{t^2+1}{t^3+1}dt=\frac{1}{2}{\left[({\left(\log (1+{\mathrm{t}}^3)\right]}_4^{\mathrm{x}}-\log (1-\mathrm{t}+{\mathrm{t}}^2))\right]}_4^{\mathrm{x}}+\frac{1}{3}{\left[\log (1+\mathrm{t})\right]}_4^{\mathrm{x}}+\frac{2}{3}\mathrm{Arc}{\left[\frac{\sqrt{3}}{3}(2\mathrm{t}-1)\right]}_4^{\mathrm{x}}$$$$=\frac{1}{2}{\left[\log \left(\frac{1+{\mathrm{t}}^3}{1-\mathrm{t}+{\mathrm{t}}^2}\right)\right]}_4^{\mathrm{x}}+\frac{1}{3}{\left[\log (1+\mathrm{t})\right]}_4^{\mathrm{x}}+\frac{2}{3}\mathrm{A}rc\tan {\left[\frac{\sqrt{3}}{3}(2t-1)\right]}_{4^{}}^x$$$$=\frac{1}{2}{\left[\log (1+\mathrm{t})\right]}_4^{\mathrm{x}}+\frac{1}{3}{\left[\log (1+\mathrm{t})\right]}_4^{\mathrm{x}}+\frac{2}{3}\mathrm{A}rc\tan {\left[\frac{\sqrt{3}}{3}(2t-1)\right]}_4^x$$$$\frac{5}{6}[\log (1+\mathrm{x})-\log 5)]+\frac{2}{3}[\mathrm{Arctan}(\frac{\sqrt{3}}{3}2\mathrm{x}-\frac{\sqrt{3}}{3})-\frac{2}{3}\mathrm{A}rc\tan \frac{7\sqrt{3}}{3}$$$$\frac{5}{6}\log \left(\frac{1+\mathrm{x}}{5}\right)+\frac{2}{3}\mathrm{A}rc\tan (\frac{2\sqrt{3}}{3}x-\frac{\sqrt{3}}{3})-\frac{2}{3}Arc\tan \left(\frac{7\sqrt{3}}{3}\right)$$$$I=2(\sqrt{x}-2)1+x=1+{\left(\sqrt{x}\right)}^2\sqrt{x}=X$$$$$$$$\log {\left[\frac{1+\mathrm{x}}{5}\right]}^{\frac{5}{6}}-2\sqrt{\mathrm{x}}+\mathrm{A}rc\tan \frac{\sqrt{3}}{3}(2x-1)=4+\frac{2}{4}Arc\tan \left(\frac{7\sqrt{3}}{3}\right)$$$$\log {\left(\frac{1+{\mathrm{X}}^2}{5}\right)}^{\frac{5}{6}}-2X+Ar\mathrm{ctan}\frac{\sqrt{3}}{3}(2X^2-1)-4-Arc\tan \left(\frac{7\sqrt{3}}{3}\right)=0$$$$...$$$$$$

Commented by Shrinava last updated on 20/Sep/22

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}\mathrm{solution}\mathrm{please}$$

Commented by a.lgnaoui last updated on 20/Sep/22

$$x=4$$

Answered by Peace last updated on 20/Sep/22

$$F\left(x\right)={\int }_4^x\frac{t^2+1}{t^3+1}dt-2(\sqrt{x}-2),DF={\mathbb{R}}_{+}$$$$F^{\prime }\left(x\right)=\frac{x^2+1}{x^3+1}-\frac{1}{\sqrt{x}},\mathrm{\forall }x\in {\mathbb{R}}_{+}^{\ast }$$$$g\left(x\right)=F^{{}^{\prime }}\left(x^2\right)=\frac{x^4+1}{x^6+1}-\frac{1}{x}=\frac{x^5+x-{\mathit{x}}^6-1}{\mathit{x}({\mathit{x}}^6+1)}=\frac{(x-1)(x^5-1)}{x(x^6+1)}$$$$\underset{[0,\mathrm{\infty }[}{x}\stackrel{h}{\to }{\underset{[0,\mathrm{\infty }[}{x}}^{2}isBijection$$$$=\frac{(x-1)(x-1)(x^4+x^3+x^2+1)}{x(x^6+1)}=\frac{{(x-1)}^2(x^4+x^3+x^2+1)}{x(x^6+1)}⩾0$$$$\mathrm{\forall }x\in {\mathbb{R}}_{+}^{\ast }$$$$\Rightarrow F\left(x\right)IsincreasingFunction$$$$F\left(4\right)=0\Rightarrow x=4istheUniQuesolution$$

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