How do you integrate (4x^2-x+3)/((x+5)(x-1)(x-2)) using partial fractions?

Aug 28, 2016

$- \ln | x - 1 | + \frac{17}{7} \ln | x - 2 | + \frac{18}{7} \ln | x + 5 | + K$, or,

$\ln | \frac{{\left(x + 5\right)}^{\frac{18}{7}} {\left(x - 2\right)}^{\frac{17}{7}}}{x - 1} | + K$.

Explanation:

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

We will use the Method of Partial Fraction to split the Integrand

(4x^2-x+3)/((x+5)(x-1)(x-2))=A/(x-1)+B/(x-2)+C/(x+5);

where, $A , B , C \in \mathbb{R}$.

To determine, $A , B , C$, let us use Heavyside's Cover-up Method :-

$A = {\left[\frac{4 {x}^{2} - x + 3}{\left(x + 5\right) \left(x - 2\right)}\right]}_{x = 1} = \frac{6}{-} 6 = - 1$;

$B = {\left[\frac{4 {x}^{2} - x + 3}{\left(x + 5\right) \left(x - 1\right)}\right]}_{x = 2} = \frac{17}{7}$;

$C = {\left[\frac{4 {x}^{2} - x + 3}{\left(x - 1\right) \left(x - 2\right)}\right]}_{x = - 5} = \frac{108}{\left(- 6\right) \left(- 7\right)} = \frac{18}{7}$.

Therefore, $I = \int \left[- \frac{1}{x - 1} + \frac{\frac{17}{7}}{x - 2} + \frac{\frac{18}{7}}{x + 5}\right] \mathrm{dx}$

$= - \ln | x - 1 | + \frac{17}{7} \ln | x - 2 | + \frac{18}{7} \ln | x + 5 | + K$, or,

$= \ln | \frac{{\left(x + 5\right)}^{\frac{18}{7}} {\left(x - 2\right)}^{\frac{17}{7}}}{x - 1} | + K$.

Enjoy Maths.!