# What is int (-3x^3+4 ) / (-2x^2+x +1 )?

Nov 22, 2017

$\int \frac{\left(- 3 {x}^{3} + 4\right) \cdot \mathrm{dx}}{- 2 {x}^{2} + x + 1}$

=$\frac{3 {x}^{2} + 3 x}{4} + \frac{35}{24} \cdot L n \left(2 x + 1\right) - \frac{1}{3} \cdot L n \left(x - 1\right) + C$

#### Explanation:

$\int \frac{\left(- 3 {x}^{3} + 4\right) \cdot \mathrm{dx}}{- 2 {x}^{2} + x + 1}$

=$\int \frac{\left(3 {x}^{3} - 4\right) \cdot \mathrm{dx}}{2 {x}^{2} - x - 1}$

=$\frac{1}{4} \cdot \int \frac{\left(12 {x}^{3} - 16\right) \cdot \mathrm{dx}}{2 {x}^{2} - x - 1}$

=$\frac{1}{4} \cdot \int \frac{\left[\left(2 {x}^{2} - x - 1\right) \cdot \left(6 x + 3\right) + 9 x - 13\right] \cdot \mathrm{dx}}{2 {x}^{2} - x - 1}$

=$\frac{1}{4} \cdot \int \left(6 x + 3\right) \cdot \mathrm{dx} + \frac{1}{4} \cdot \int \frac{\left(9 x - 13\right) \cdot \mathrm{dx}}{2 {x}^{2} - x - 1}$

=$\frac{3 {x}^{2} + 3 x}{4}$+$\frac{1}{4} \cdot \int \frac{\left(9 x - 13\right) \mathrm{dx}}{\left(2 x + 1\right) \left(x - 1\right)}$

=$\frac{3 {x}^{2} + 3 x}{4} + \frac{1}{4} \cdot \left[\frac{35}{3} \cdot \int \frac{\mathrm{dx}}{2 x + 1} - \frac{4}{3} \cdot \int \frac{\mathrm{dx}}{x - 1}\right]$

=$\frac{3 {x}^{2} + 3 x}{4} + \frac{35}{12} \cdot \int \frac{\mathrm{dx}}{2 x + 1} - \frac{1}{3} \cdot \int \frac{\mathrm{dx}}{x - 1}$

=$\frac{3 {x}^{2} + 3 x}{4} + \frac{35}{24} \cdot L n \left(2 x + 1\right) - \frac{1}{3} \cdot L n \left(x - 1\right) + C$