What is int (-2x^3-x^2+x+2 ) / (3x^2+2x +3 )?

Dec 17, 2017

$\frac{25 L n \left(3 {x}^{2} + 2 x + 3\right) + 10 \sqrt{2} \arctan \left(\frac{3 x + 1}{2 \sqrt{2}}\right) - 18 {x}^{2} + 6 x}{54} + C$ or

=$\frac{25}{54} L n \left(3 {x}^{2} + 2 x + 3\right) + \frac{5 \sqrt{2}}{27} \arctan \left(\frac{3 x + 1}{2 \sqrt{2}}\right) - \frac{3 {x}^{2} - x}{9} + C$

Explanation:

$\int \frac{- 2 {x}^{3} - {x}^{2} + x + 2}{3 {x}^{2} + 2 x + 3} \cdot \mathrm{dx}$

=$- \frac{1}{9} \int \frac{18 {x}^{3} + 9 {x}^{2} - 9 x - 18}{3 {x}^{2} + 2 x + 3} \cdot \mathrm{dx}$

=$- \frac{1}{9} \int \frac{\left(6 x - 1\right) \cdot \left(3 {x}^{2} + 2 x + 3\right) - \left(25 x + 15\right)}{3 {x}^{2} + 2 x + 3} \cdot \mathrm{dx}$

=$- \frac{1}{9} \int \left(6 x - 1\right) \mathrm{dx}$+$\frac{1}{9} \int \frac{25 x + 15}{3 {x}^{2} + 2 x + 3} \cdot \mathrm{dx}$

=$- \frac{1}{9} \cdot \left(3 {x}^{2} - x\right) + C$+$\frac{1}{9} \int \frac{\frac{25}{6} \cdot \left(6 x + 2\right) + \frac{20}{3}}{3 {x}^{2} + 2 x + 3} \cdot \mathrm{dx}$

=$- \frac{1}{9} \cdot \left(3 {x}^{2} - x\right) + C$+$\frac{25}{54} \int \frac{6 x + 2}{3 {x}^{2} + 2 x + 3} \cdot \mathrm{dx}$+$\frac{20}{27} \int \frac{\mathrm{dx}}{3 {x}^{2} + 2 x + 3}$

=$- \frac{1}{9} \cdot \left(3 {x}^{2} - x\right) + \frac{25}{54} \ln \left(3 {x}^{2} + 2 x + 3\right) + C$+$\frac{20}{27} \int \frac{3 \mathrm{dx}}{9 {x}^{2} + 6 x + 9}$

=$- \frac{1}{9} \cdot \left(3 {x}^{2} - x\right) + \frac{25}{54} \ln \left(3 {x}^{2} + 2 x + 3\right) + C$+$\frac{20}{27} \int \frac{3 \mathrm{dx}}{{\left(3 x + 1\right)}^{2} + {\left(2 \sqrt{2}\right)}^{2}}$

=$- \frac{1}{9} \cdot \left(3 {x}^{2} - x\right) + \frac{25}{54} \ln \left(3 {x}^{2} + 2 x + 3\right)$+$\frac{5 \sqrt{2}}{27} \arctan \left(\frac{3 x + 1}{2 \sqrt{2}}\right) + C$

=$\frac{25 L n \left(3 {x}^{2} + 2 x + 3\right) + 10 \sqrt{2} \arctan \left(\frac{3 x + 1}{2 \sqrt{2}}\right) - 18 {x}^{2} + 6 x}{54} + C$