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Question Number 87503 by M±th+et£s last updated on 04/Apr/20

$$\int \frac{x^2}{1+x^5}dx$$

Commented by MJS last updated on 04/Apr/20

$$x^5+1=(x+1)(x^2-\frac{1-\sqrt{5}}{2}x+1)(x^2-\frac{1+\sqrt{5}}{2}+1)$$$$\mathrm{can}\mathrm{you}\mathrm{do}\mathrm{it}\mathrm{now}?$$

Commented by M±th+et£s last updated on 06/Apr/20

$$siritryedalotbutthereisnosolution$$$$howcanyoudoit$$

Commented by MJS last updated on 06/Apr/20

$$\mathrm{I}\mathrm{will}\mathrm{show}\mathrm{later}$$

Commented by M±th+et£s last updated on 06/Apr/20

$$thankyousir$$

Answered by MJS last updated on 06/Apr/20

$$x^5+1=0$$$$x_1=-1$$$$x_{2,3}={\mathrm{e}}^{\pm \mathrm{i}\frac{\pi }{5}}$$$$x_{4,5}={\mathrm{e}}^{\pm \mathrm{i}\frac{3\pi }{5}}$$$$x^5+1=(x-x_1)\left((x-x_2)(x-x_3)\right)\left((x-x_4)(x-x_5)\right)=$$$$=(x+1)(x^2-\frac{1+\sqrt{5}}{2}x+1)(x^2-\frac{1-\sqrt{5}}{2}x+1)$$$$\Rightarrow$$$$\frac{x^2}{x^5+1}=\frac{\alpha }{x+1}+\frac{\beta x+\gamma }{x^2-\frac{1+\sqrt{5}}{2}x+1}+\frac{\delta x+ϵ}{x^2-\frac{1-\sqrt{5}}{2}x+1}$$$$\Rightarrow$$$$\alpha =\frac{1}{5}$$$$\beta =\gamma =-\frac{1}{10}+\frac{\sqrt{5}}{10}$$$$\delta =ϵ=-\frac{1}{10}-\frac{\sqrt{5}}{10}$$$$\Rightarrow$$$$\int \frac{x^2}{x^5+1}dx=$$$$=\frac{1}{5}\int \frac{dx}{x+1}-\frac{1-\sqrt{5}}{10}\int \frac{x+1}{x^2-\frac{1+\sqrt{5}}{2}x+1}dx-\frac{1+\sqrt{5}}{10}\int \frac{x+1}{x^2-\frac{1-\sqrt{5}}{2}x+1}dx$$$$\mathrm{now}\mathrm{use}\mathrm{formulas}$$

Commented by M±th+et£s last updated on 06/Apr/20

$$godblessyousir$$

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