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Question Number 64662 by aliesam last updated on 20/Jul/19

$$\int \frac{dx}{{(x^8+x^4+1)}^2}$$$$$$$${\int }_{\frac{1}{x}}^x\frac{ln\left(t\right)}{t^2+1}dt$$

Commented by MJS last updated on 20/Jul/19

$$\mathrm{the}{2}^{\mathrm{nd}}\mathrm{one}\mathrm{should}\mathrm{be}=0$$

Commented by aliesam last updated on 20/Jul/19

$$yesthatsright$$

Commented by mathmax by abdo last updated on 20/Jul/19

$$2)letI={\int }_{\frac{1}{x}}^x\frac{ln\left(t\right)}{1+t^2}dtwithx>0changementt=\frac{1}{u}give$$$$I=-{\int }_{\frac{1}{x}}^x\frac{-lnu}{1+\frac{1}{u^2}}(-\frac{du}{u^2})=-{\int }_{\frac{1}{x}}^x\frac{lnu}{u^2+1}=-I\Rightarrow 2I=0\Rightarrow I=0$$$$remarkwhenxgoto+\mathrm{\infty }weget{\int }_0^{+\mathrm{\infty }}\frac{lnt}{t^2+1}dt=0$$

Answered by MJS last updated on 20/Jul/19

$$\int \frac{dx}{{(x^8+x^4+1)}^2}$$$${\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}$$$$\int \frac{P\left(x\right)}{Q\left(x\right)}dx=\frac{P_1\left(x\right)}{Q_1\left(x\right)}+\int \frac{P_2\left(x\right)}{Q_2\left(x\right)}dx$$$$Q_1\left(x\right)=\gcd (Q\left(x\right),Q^{\prime }\left(x\right))$$$$Q_2\left(x\right)=\frac{Q\left(x\right)}{Q_1\left(x\right)}$$$$P_1\left(x\right),P_2\left(x\right)\mathrm{can}\mathrm{be}\mathrm{found}\mathrm{by}\mathrm{comparing}\mathrm{the}$$$$\mathrm{factors}\mathrm{of}\frac{P\left(x\right)}{Q\left(x\right)}={\left(\frac{P_1\left(x\right)}{Q_1\left(x\right)}\right)}^{\prime }+\frac{P_2\left(x\right)}{Q_2\left(x\right)}$$$$\mathrm{in}\mathrm{our}\mathrm{case}$$$$P\left(x\right)=1Q\left(x\right)={(x^8+x^4+1)}^2$$$$Q_1\left(x\right)=Q_2\left(x\right)=x^8+x^4+1$$$$P_1\left(x\right)=\frac{1}{12}(x-x^5)P_2\left(x\right)=\frac{1}{12}(11-3x^4)$$$$\int \frac{dx}{{(x^8+x^4+1)}^2}=-\frac{x^5-x}{12(x^8+x^4+1)}-\frac{1}{12}\int \frac{3x^4-11}{x^8+x^4+1}dx$$$$$$$$\frac{3x^4-11}{x^8+x^4+1}=\frac{3x^4-11}{(x^2-x+1)(x^2+x+1)(x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)}=$$$$=-\frac{3x+11}{4(x^2-x+1)}+\frac{3x-11}{4(x^2+x+1)}+\frac{25x-11\sqrt{3}}{4\sqrt{3}(x^2-\sqrt{3}x+1)}-\frac{25x+11\sqrt{3}}{4\sqrt{3}(x^2+\sqrt{3}x+1)}$$$$$$$$\mathrm{now}\mathrm{use}\mathrm{this}\mathrm{formula}$$$$\int \frac{ax+b}{x^2+cx+d}dx=\frac{a}{2}\int \frac{2x+c}{x^2+cx+d}dx+\frac{2b-ac}{2}\int \frac{dx}{x^2+cx+d}=$$$$=\frac{a}{2}\ln (x^2+cx+d)-\frac{ac-2b}{\sqrt{4d-c^2}}\arctan \frac{2x+c}{\sqrt{4d-c^2}}$$

Commented by aliesam last updated on 20/Jul/19

$$godblessuousir$$

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