# What is int (-2x^3-2x^2+6x+9 ) / (2x^2- x +3 )?

Aug 2, 2018

$I = - {x}^{2} / 2 - \frac{3}{2} x + \frac{25 \sqrt{23}}{32} \ln \left(| \frac{16}{23} {\left(x - \frac{1}{4}\right)}^{2} + 1 |\right) + \frac{29}{24} \frac{\sqrt{23}}{92} {\tan}^{- 1} \left(\frac{4}{\sqrt{23}} \left(x - \frac{1}{4}\right)\right) + C$, $C \in \mathbb{R}$

#### Explanation:

$I = \int \frac{- 2 {x}^{3} - 2 {x}^{2} + 6 x + 9}{2 {x}^{2} - x + 3} \mathrm{dx}$

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

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

$= - \int \frac{x \left({x}^{2} - \frac{1}{2} x + \frac{3}{2}\right) + \frac{3}{2} \left({x}^{2} - \frac{1}{2} x + \frac{3}{2}\right) - \frac{15}{4} x - \frac{27}{4}}{{x}^{2} - \frac{1}{2} x + \frac{3}{2}} \mathrm{dx}$

$= - \int \left(x + \frac{3}{2} - \frac{3}{4} \cdot \frac{5 x + 9}{{x}^{2} - \frac{1}{2} x + \frac{3}{2}}\right) \mathrm{dx}$

$= - \int x \mathrm{dx} - \frac{3}{2} \int 1 \mathrm{dx} + \frac{15}{4} \int \frac{x}{{\left(x - \frac{1}{4}\right)}^{2} + \frac{23}{16}} \mathrm{dx} + \frac{27}{4} \int \frac{1}{{\left(x - \frac{1}{4}\right)}^{2} + \frac{23}{16}} \mathrm{dx}$

Let $x - \frac{1}{4} = \sqrt{\frac{23}{16}} \tan \left(\theta\right)$

$\mathrm{dx} = \sqrt{\frac{23}{16}} \sec {\left(\theta\right)}^{2} d \theta$

So :

$I = - {x}^{2} / 2 - \frac{3}{2} x + \frac{60}{23} \int \frac{\sqrt{\frac{23}{16}} \tan \left(\theta\right) + \frac{1}{4}}{\left(\tan {\left(\theta\right)}^{2} + 1\right)} \cdot \sqrt{\frac{23}{16}} \sec {\left(\theta\right)}^{2} d \theta + \frac{101}{23} \int \frac{1}{\tan {\left(\theta\right)}^{2} + 1} \cdot \sqrt{\frac{23}{16}} \sec {\left(\theta\right)}^{2} d \theta$

Because $\tan {\left(\theta\right)}^{2} + 1 = \sec {\left(\theta\right)}^{2}$,
$= - {x}^{2} / 2 - \frac{3}{2} x + \frac{60}{23} \sqrt{\frac{23}{16}} \int \left(\frac{23}{16} \tan \left(\theta\right) + \frac{1}{4}\right) d \theta + \frac{101}{23} \sqrt{\frac{23}{16}} \int 1 d \theta$

$= - {x}^{2} / 2 - \frac{3}{2} x + \frac{15}{4} \sqrt{\frac{23}{16}} \int \tan \left(\theta\right) d \theta + \frac{15}{23} \sqrt{\frac{23}{16}} \int 1 d \theta + \frac{101}{23} \sqrt{\frac{23}{16}} \theta$

We have $\int \tan \left(\theta\right) d \theta = - \ln \left(| \cos \left(\theta\right) |\right)$, you can see the proof here.

$= - {x}^{2} / 2 - \frac{3}{2} x - \frac{25 \sqrt{23}}{16} \ln \left(| \cos \left(\theta\right) |\right) + \frac{15 \sqrt{23}}{92} \theta + \frac{101 \sqrt{23}}{92} \theta$

Because $\theta = {\tan}^{- 1} \left(\frac{4}{\sqrt{23}} \left(x - \frac{1}{4}\right)\right)$, and $\cos \left(A r c \tan \left(x\right)\right) = \frac{1}{\sqrt{{x}^{2} + 1}}$ (you can see the proof here),

$I = - {x}^{2} / 2 - \frac{3}{2} x + \frac{25 \sqrt{23}}{32} \ln \left(| \frac{16}{23} {\left(x - \frac{1}{4}\right)}^{2} + 1 |\right) + \frac{29}{24} \frac{\sqrt{23}}{92} {\tan}^{- 1} \left(\frac{4}{\sqrt{23}} \left(x - \frac{1}{4}\right)\right) + C$, $C \in \mathbb{R}$