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Question Number 76910 by aliesam last updated on 31/Dec/19

Commented by MJS last updated on 31/Dec/19

$$x^6+1=(x^2+1)(x^2+\sqrt{3}x+1)(x^2-\sqrt{3}x+1)$$

Answered by MJS last updated on 01/Jan/20

$$\int \frac{dx}{x^6+1}=$$$$=\frac{1}{3}\int \frac{dx}{x^2+1}-\frac{1}{6}\int \frac{\sqrt{3}x-2}{x^2-\sqrt{3}x+1}dx+\frac{1}{6}\int \frac{\sqrt{3}x+2}{x^2+\sqrt{3}x+1}dx$$$$I_1=\frac{1}{3}\int \frac{dx}{x^2+1}=\frac{1}{3}\arctan x$$$$I_2=-\frac{1}{6}\int \frac{\sqrt{3}x-2}{x^2-\sqrt{3}x+1}dx=$$$$=-\frac{\sqrt{3}}{12}\int \frac{2x-\sqrt{3}}{x^2-\sqrt{3}x+1}dx+\frac{1}{12}\int \frac{dx}{x^2-\sqrt{3}x+1}=$$$$=-\frac{\sqrt{3}}{12}\ln (x^2-\sqrt{3}x+1)+\frac{1}{6}\arctan (2x-\sqrt{3})$$$$I_3=\frac{1}{6}\int \frac{\sqrt{3}x+2}{x^2+\sqrt{3}x+1}dx=$$$$=\frac{\sqrt{3}}{12}\int \frac{2x+\sqrt{3}}{x^2+\sqrt{3}x+1}dx+\frac{1}{12}\int \frac{dx}{x^2+\sqrt{3}x+1}=$$$$=\frac{\sqrt{3}}{12}\ln (x^2+\sqrt{3}x+1)+\frac{1}{6}\arctan (2x+\sqrt{3})$$$$I=I_1+I_2+I_3+C$$

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