# How do you integrate int (3x^3 - 22x^2 + 27x + 46)/(x^2 - 8x + 15) using partial fractions?

Oct 2, 2016

$\int \frac{3 {x}^{3} - 22 {x}^{2} + 27 x + 46}{{x}^{2} - 8 x + 15} \mathrm{dx} = \frac{3}{2} {x}^{2} + 2 x - 5 \ln \left\mid x - 3 \right\mid + 3 \ln \left\mid x - 5 \right\mid + C$

#### Explanation:

$\frac{3 {x}^{3} - 22 {x}^{2} + 27 x + 46}{{x}^{2} - 8 x + 15} = \frac{\left(3 {x}^{3} - 24 {x}^{2} + 45 x\right) + \left(2 {x}^{2} - 16 x + 30\right) + \left(- 2 x + 16\right)}{{x}^{2} - 8 x + 15}$

$\textcolor{w h i t e}{\frac{3 {x}^{3} - 22 {x}^{2} + 27 x + 46}{{x}^{2} - 8 x + 15}} = 3 x + 2 + \frac{- 2 x + 16}{{x}^{2} - 8 x + 15}$

$\textcolor{w h i t e}{\frac{3 {x}^{3} - 22 {x}^{2} + 27 x + 46}{{x}^{2} - 8 x + 15}} = 3 x + 2 + \frac{- 2 x + 16}{\left(x - 3\right) \left(x - 5\right)}$

$\textcolor{w h i t e}{\frac{3 {x}^{3} - 22 {x}^{2} + 27 x + 46}{{x}^{2} - 8 x + 15}} = 3 x + 2 + \frac{A}{x - 3} + \frac{B}{x - 5}$

We can determine $A$ and $B$ using Heaviside's cover up method...

$A = \frac{- 2 \left(\textcolor{b l u e}{3}\right) + 16}{\textcolor{b l u e}{3} - 5} = \frac{10}{- 2} = - 5$

$B = \frac{- 2 \left(\textcolor{b l u e}{5}\right) + 16}{\textcolor{b l u e}{5} - 3} = \frac{6}{2} = 3$

So:

$\int \frac{3 {x}^{3} - 22 {x}^{2} + 27 x + 46}{{x}^{2} - 8 x + 15} \mathrm{dx} = \int 3 x + 2 - \frac{5}{x - 3} + \frac{3}{x - 5} \mathrm{dx}$

$\textcolor{w h i t e}{\int \frac{3 {x}^{3} - 22 {x}^{2} + 27 x + 46}{{x}^{2} - 8 x + 15} \mathrm{dx}} = \frac{3}{2} {x}^{2} + 2 x - 5 \ln \left\mid x - 3 \right\mid + 3 \ln \left\mid x - 5 \right\mid + C$