# How do you integrate int (x^2 - 8x + 44) / ((x + 2) (x - 2)^2) using partial fractions?

Jul 11, 2017

I decomposed integrand into basic fractions,

$\frac{{x}^{2} - 8 x + 44}{\left(x + 2\right) \cdot {\left(x - 2\right)}^{2}} = \frac{A}{x + 2} + \frac{B}{x - 2} + \frac{C}{x - 2} ^ 2$

$\left({x}^{2} - 8 x + 44\right) = A \cdot {\left(x - 2\right)}^{2} + B \cdot \left({x}^{2} - 4\right) + C \cdot \left(x + 2\right)$

$\left({x}^{2} - 8 x + 44\right) = A \cdot \left({x}^{2} - 4 x + 4\right) + B \cdot \left({x}^{2} - 4\right) + C \cdot \left(x + 2\right)$

$\left({x}^{2} - 8 x + 44\right) = \left(A + B\right) \cdot {x}^{2} + \left(- 4 A + C\right) \cdot x + \left(4 A - 4 B + 2 C\right)$

After equating coefficients, I found $A + B = 1$, $- 4 A + C = - 8$ and $4 A - 4 B + 2 C = 44$ equations,

After solving system of them simultaneously, I found;

$A = 4 , B = - 3$ and $C = 8$.

Thus,

$\int \frac{{x}^{2} - 8 x + 44}{\left(x + 2\right) \cdot {\left(x - 2\right)}^{2}} \mathrm{dx}$

=$\int 4 \frac{\mathrm{dx}}{x + 2} - \int 3 \frac{\mathrm{dx}}{x - 2} + \int 8 \frac{\mathrm{dx}}{x - 2} ^ 2$

=$4 L n \left(| x + 2 |\right) - 3 L n \left(| x - 2 |\right) - \frac{8}{x - 2} + C$

#### Explanation:

I decomposed integrand into basic fractions.