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

May 23, 2017

${x}^{2} / 2 + \frac{x}{2} + \frac{17}{8} \ln | - 2 {x}^{2} + x + 7 | + \frac{45}{8 \sqrt{57}} \ln | \frac{4 x - 1 - \sqrt{57}}{4 x - 1 + \sqrt{57}} | + C .$

#### Explanation:

Let, $I = \int \frac{- 2 {x}^{3} - x}{- 2 {x}^{2} + x + 7} \mathrm{dx}$

We have, $- 2 {x}^{3} - x = x \left(- 2 {x}^{2} + x + 7\right) + \frac{1}{2} \left(- 2 {x}^{2} + x + 7\right) - \frac{17}{2} x - \frac{7}{2.}$

$\therefore \frac{- 2 {x}^{3} - x}{- 2 {x}^{2} + x + 7} = x + \frac{1}{2} + \frac{- 17 \frac{x}{2} - \frac{7}{2}}{- 2 {x}^{2} + x + 7} .$

Also, $\frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} + x + 7\right) = - 4 x + 1 , \text{ we set, } - \frac{17}{2} x - \frac{7}{2} = \frac{17}{8} \left(- 4 x + 1\right) - \frac{45}{8.}$

Thus, we have,

$\frac{- 2 {x}^{3} - x}{- 2 {x}^{2} + x + 7} = x + \frac{1}{2} + \frac{\frac{17}{8} \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} + x + 7\right) - \frac{45}{8}}{- 2 {x}^{2} + x + 7} .$

$\therefore I = \int \left[x + \frac{1}{2} + \frac{\frac{17}{8} \frac{d}{\mathrm{dx}} \left(- 2 {x}^{2} + x + 7\right) - \frac{45}{8}}{- 2 {x}^{2} + x + 7}\right] \mathrm{dx} ,$

$= {x}^{2} / 2 + \frac{x}{2} + \frac{17}{8} \ln | - 2 {x}^{2} + x + 7 | + \frac{45}{8} \int \frac{1}{2 {x}^{2} - x - 7} \mathrm{dx} ,$

$= {x}^{2} / 2 + \frac{x}{2} + \frac{17}{8} \ln | - 2 {x}^{2} + x + 7 | + \frac{45}{16} \int \frac{1}{{x}^{2} - \frac{x}{2} - \frac{7}{2}} \mathrm{dx} ,$

$= {x}^{2} / 2 + \frac{x}{2} + \frac{17}{8} \ln | - 2 {x}^{2} + x + 7 | + \frac{45}{16} \int \frac{\mathrm{dx}}{{\left(x - \frac{1}{4}\right)}^{2} - \frac{57}{16}} ,$

$= {x}^{2} / 2 + \frac{x}{2} + \frac{17}{8} \ln | - 2 {x}^{2} + x + 7 |$

$+ \frac{45}{16} \cdot \frac{1}{2 \left(\frac{\sqrt{57}}{4}\right)} \ln | \frac{x - \frac{1}{4} - \frac{\sqrt{57}}{4}}{x - \frac{1}{4} + \frac{\sqrt{57}}{4}} | ,$

$\Rightarrow I = {x}^{2} / 2 + \frac{x}{2} + \frac{17}{8} \ln | - 2 {x}^{2} + x + 7 |$

$+ \frac{45}{8 \sqrt{57}} \ln | \frac{4 x - 1 - \sqrt{57}}{4 x - 1 + \sqrt{57}} | + C .$

Enjoy Maths.!