# How do you integrate (3x^3-5x^2-11x+9)/(x^2-2x-3) using partial fractions?

Jul 31, 2017

$\int \frac{3 {x}^{3} - 5 {x}^{2} - 11 x + 9}{{x}^{2} - 2 x - 3} \mathrm{dx}$

= $\frac{3 {x}^{2}}{2} + x + 3 \ln | x - 3 | - 3 \ln | x + 1 | + c$

#### Explanation:

For converting to partial fractions, we must have degreeof numerator less than that of denominator. Hence as

$3 {x}^{3} - 5 {x}^{2} - 11 x + 9 = 3 x \left({x}^{2} - 2 x - 3\right) + 1 \left({x}^{2} - 2 x - 3\right) + 12$ and hence

$\frac{3 {x}^{3} - 5 {x}^{2} - 11 x + 9}{{x}^{2} - 2 x - 3} = 3 x + 1 + \frac{12}{{x}^{2} - 2 x - 3}$

As $\left({x}^{2} - 2 x - 3\right) = \left(x - 3\right) \left(x + 1\right)$, let

$\frac{12}{{x}^{2} - 2 x - 3} \Leftrightarrow \frac{A}{x - 3} + \frac{B}{x + 1} = \frac{A \left(x + 1\right) + B \left(x - 3\right)}{{x}^{2} - 2 x - 3}$

if we put $x = 3$, $4 A = 12$ i.e. $A = 3$

and if we put $x = - 1$, $- 4 B = 12$ or $B = - 3$

Hence $\frac{3 {x}^{3} - 5 {x}^{2} - 11 x + 9}{{x}^{2} - 2 x - 3} = 3 x + 1 + \frac{12}{{x}^{2} - 2 x - 3}$

and $\int \frac{3 {x}^{3} - 5 {x}^{2} - 11 x + 9}{{x}^{2} - 2 x - 3} \mathrm{dx}$

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

= $\frac{3 {x}^{2}}{2} + x + 3 \ln | x - 3 | - 3 \ln | x + 1 | + c$