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Question Number 73477 by abdomathmax last updated on 13/Nov/19

$$calculate\int \frac{x^3-4x+5}{x^2-x+1}dx$$

Answered by MJS last updated on 13/Nov/19

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

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