# How do you integrate int (3x^2 - 1) /( ((x^2)+2) (x-3)) using partial fractions?

Aug 17, 2016

$\frac{26}{11} \ln | x - 3 | + \frac{7}{22} \ln \left({x}^{2} + 2\right) + \frac{21}{11 \sqrt{2}} {\tan}^{- 1} \left(\frac{x}{\sqrt{2}}\right) + K$.

#### Explanation:

We split the Integrand as :
$\frac{3 {x}^{2} - 1}{\left({x}^{2} + 2\right) \left(x - 3\right)} = \frac{A}{x - 3} + \frac{B x + C}{{x}^{2} + 2} , \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(\star\right)$
where, $A , B , C \in \mathbb{R}$.

$A$ can be quickly determined by Heavyside"s Cover-up Method as

$A = {\left[\frac{3 {x}^{2} - 1}{{x}^{2} + 2}\right]}_{x = 3} = \frac{26}{11}$

Simplifying $\left(\star\right)$ & comparing rational polys. on both sides, we get,
$A \left({x}^{2} + 2\right) + \left(B x + C\right) \left(x - 3\right) = 3 {x}^{2} - 1. \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$

Taking x=0, &, A=26/11 in $\left(1\right)$,
$\frac{26}{11} \left(2\right) + C \left(- 3\right) = - 1 \Rightarrow C = \frac{21}{11}$.

$\left(1\right) , x = 1 , A = \frac{26}{11} , C = \frac{21}{11}$
$\Rightarrow \frac{26}{11} \left(3\right) + \left(B + \frac{21}{11}\right) \left(- 2\right) = 2 \Rightarrow B = \frac{7}{11}$.

With these $A , B , C$, we have, $\int \frac{3 {x}^{2} - 1}{\left({x}^{2} + 2\right) \left(x - 3\right)} \mathrm{dx}$,

$= \frac{26}{11} \int \frac{1}{x - 3} \mathrm{dx} + \frac{1}{11} \int \frac{7 x + 21}{{x}^{2} + 2} \mathrm{dx}$,

$= \frac{26}{11} \ln | x - 3 | + \frac{7}{11} \cdot \frac{1}{2} \int \frac{2 x}{{x}^{2} + 2} \mathrm{dx} + \frac{21}{11} \int \frac{1}{{x}^{2} + 2} \mathrm{dx}$,

$= \frac{26}{11} \ln | x - 3 | + \frac{7}{22} \int \frac{d \left({x}^{2} + 2\right)}{{x}^{2} + 2} + \frac{21}{11} \cdot \frac{1}{\sqrt{2}} \cdot {\tan}^{- 1} \left(\frac{x}{\sqrt{2}}\right)$,

$= \frac{26}{11} \ln | x - 3 | + \frac{7}{22} \ln \left({x}^{2} + 2\right) + \frac{21}{11 \sqrt{2}} {\tan}^{- 1} \left(\frac{x}{\sqrt{2}}\right) + K$.

Enjoy Maths!