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Question Number 86853 by M±th+et£s last updated on 01/Apr/20

$$\int \frac{x^6-x^3+2}{x^4-x^2-2}dx$$

Answered by MJS last updated on 01/Apr/20

$$\frac{x^6-x^3+2}{x^4-x^2-2}=$$$$=x^2+1-\frac{x^3-3x^2-4}{x^4-x^2-2}$$$$\frac{x^3-3x^2-4}{x^4-x^2-2}=$$$$=-(\frac{x}{3(x^2+1)}+\frac{1}{3(x^2+1)}+\frac{2x}{3(x^2-2)}-\frac{5\sqrt{2}}{6(x-\sqrt{2})}+\frac{5\sqrt{2}}{6(x+\sqrt{2})})$$$$\int \frac{x^6-x^3+2}{x^4-x^2-2}dx=$$$$=\int x^{2}dx+\int dx-\frac{1}{6}\int \frac{2x}{x^2+1}dx-\frac{1}{3}\int \frac{dx}{x^2+1}-\frac{1}{3}\int \frac{2x}{x^2-2}dx+\frac{5\sqrt{2}}{6}\int \frac{dx}{x-\sqrt{2}}-\frac{5\sqrt{2}}{6}\int \frac{dx}{x+\sqrt{2}}=$$$$=\frac{1}{3}{x}^3+x-\frac{1}{6}\ln (x^2+1)-\frac{1}{3}\arctan x-\frac{1}{3}\ln (x^2-2)+\frac{5\sqrt{2}}{6}\ln (x-\sqrt{2})-\frac{5\sqrt{2}}{6}\ln (x+\sqrt{2})=$$$$=\frac{x^3+3x}{3}-\frac{1}{6}\ln \left((x^2+1){(x^2-2)}^2\right)+\frac{5\sqrt{2}}{6}\ln \mid \frac{x-\sqrt{2}}{x+\sqrt{2}}\mid -\frac{1}{3}\arctan x+C$$

Commented by M±th+et£s last updated on 01/Apr/20

$$thankyousir$$

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