# What is int (-2x^3+4x ) / (-2x^2+x +7 )?

Oct 15, 2016

$\int \frac{2 {x}^{3} - 4 x}{2 {x}^{2} - x - 7} \mathrm{dx} = {x}^{2} / 2 + \frac{x}{2} + \frac{7}{456} \left(\left(57 + 5 \sqrt{57}\right) \ln \left(- 4 x + \sqrt{57} + 1\right) + \left(57 - 5 \sqrt{57}\right) \ln \left(4 x + \sqrt{57} - 1\right)\right) + C$

#### Explanation:

$\int \frac{- 2 {x}^{3} + 4 x}{- 2 {x}^{2} + x + 7} \mathrm{dx}$

Multiply by 1 in the form $\frac{- 1}{-} 1$:

$\int \frac{2 {x}^{3} - 4 x}{2 {x}^{2} - x - 7} \mathrm{dx}$

The power of the numeration is greater than the power of the denominator, therefore, we do the implied division:

$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots x + \frac{1}{2}$
$2 {x}^{2} - x - 7 | 2 {x}^{3} + 0 {x}^{2} - 4 x + 0$
$\ldots \ldots \ldots \ldots \ldots \ldots - 2 {x}^{2} + {x}^{2} + 7 x$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . {x}^{2} + 3 x$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots - {x}^{2} + \frac{1}{2} x + \frac{7}{2}$
$\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(\frac{7}{2}\right) x + \frac{7}{2}$

$\int \frac{2 {x}^{3} - 4 x}{2 {x}^{2} - x - 7} \mathrm{dx} = \int x \mathrm{dx} + \frac{1}{2} \int \mathrm{dx} + \frac{7}{2} \int \frac{x + 1}{2 {x}^{2} - x - 7} \mathrm{dx}$

$\int x \mathrm{dx} = {x}^{2} / 2$:

$\int \frac{2 {x}^{3} - 4 x}{2 {x}^{2} - x - 7} \mathrm{dx} = {x}^{2} / 2 + \frac{1}{2} \int \mathrm{dx} + \frac{7}{2} \int \frac{x + 1}{2 {x}^{2} - x - 7} \mathrm{dx}$

$\frac{1}{2} \int \mathrm{dx} = \frac{x}{2}$:

$\int \frac{2 {x}^{3} - 4 x}{2 {x}^{2} - x - 7} \mathrm{dx} = {x}^{2} / 2 + \frac{x}{2} + \frac{7}{2} \int \frac{x + 1}{2 {x}^{2} - x - 7} \mathrm{dx}$

$\int \frac{2 {x}^{3} - 4 x}{2 {x}^{2} - x - 7} \mathrm{dx} = {x}^{2} / 2 + \frac{x}{2} + \frac{7}{456} \left(\left(57 + 5 \sqrt{57}\right) \ln \left(- 4 x + \sqrt{57} + 1\right) + \left(57 - 5 \sqrt{57}\right) \ln \left(4 x + \sqrt{57} - 1\right)\right) + C$