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Question Number 28701 by students last updated on 29/Jan/18

$$integration\frac{x^3}{x^2+x+1}$$

Commented by abdo imad last updated on 29/Jan/18

$$\int \frac{x^3}{x^2+x+1}dx=\int \frac{x^3-1+1}{x^2+x+1}dx=\int (x-1)dx++\int \frac{dx}{x^2+x+1}$$$$=\frac{x^2}{2}-x+\int \frac{dx}{x^2+x+1}but$$$$I=\int \frac{dx}{x^2+x+1}=\int \frac{dx}{{(x+\frac{1}{2})}^2+\frac{3}{4}}letusethech.x+\frac{1}{2}=\frac{\sqrt{3}}{2}t$$$$I=\int \frac{1}{\frac{3}{4}{t}^2+\frac{3}{4}}\frac{\sqrt{3}}{2}dt=\frac{\sqrt{3}}{2}\frac{4}{3}\int \frac{dt}{1+t^2}=\frac{2\sqrt{3}}{3}arctant$$$$=\frac{2\sqrt{3}}{3}arctan\left(\frac{2}{\sqrt{3}}(x+\frac{1}{2})\right)so$$$$\int \frac{x^3}{x^2+x+1}dx=\frac{x^2}{2}-x+\frac{2\sqrt{3}}{3}arctan\left(\frac{2}{\sqrt{3}}(x+\frac{1}{2})\right).$$$$$$

Commented by abdo imad last updated on 29/Jan/18

$$\int (...)dx=........+\lambda .$$

Answered by mrW2 last updated on 29/Jan/18

$$=\frac{x^3+x^2+x-x^2-x-1+1}{x^2+x+1}$$$$=x-1+\frac{1}{x^2+x+1}$$$$=x-1+\frac{1}{{(x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$$$$\int \frac{x^3}{x^2+x+1}dx$$$$=\frac{x^2}{2}-x+\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{x+\frac{1}{2}}{\frac{\sqrt{3}}{2}}+C$$$$=\frac{x^2}{2}-x+\frac{2}{\sqrt{3}}{\tan }^{-1}\frac{2x+1}{\sqrt{3}}+C$$

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