# How do you integrate int (3(4y²-7y-12))/(y(y+2)(y-3)) using partial fractions?

Jan 13, 2017

The answer is =6ln(∣y∣)+27/5ln(∣y+2∣)+3/5ln(∣y-3∣)+C

#### Explanation:

Perform the decomposition into partial fractions

$\frac{3 \left(4 {y}^{2} - 7 y - 12\right)}{y \left(y + 2\right) \left(y - 3\right)} = \frac{A}{y} + \frac{B}{y + 2} + \frac{C}{y - 3}$

$= \frac{A \left(y + 2\right) \left(y - 3\right) + B \left(y\right) \left(y - 3\right) + C \left(y\right) \left(y + 2\right)}{y \left(y + 2\right) \left(y - 3\right)}$

Therefore, equalising the numerators

$12 {y}^{2} - 21 y - 36 = A \left(y + 2\right) \left(y - 3\right) + B \left(y\right) \left(y - 3\right) + C \left(y\right) \left(y + 2\right)$

Let $y = 0$, $\implies$, $- 36 = - 6 A$, $\implies$, $A = 6$

Let $y = - 2$,$\implies$, $48 + 42 - 36 = 10 B$, $\implies$, $B = \frac{54}{10} = \frac{27}{5}$

Let $y = 3$, $\implies$, $108 - 63 - 36 = 15 C$, $\implies$, $C = \frac{9}{15} = \frac{3}{5}$

So,

$\frac{3 \left(4 {y}^{2} - 7 y - 12\right)}{y \left(y + 2\right) \left(y - 3\right)} = \frac{6}{y} + \frac{\frac{27}{5}}{y + 2} + \frac{\frac{3}{5}}{y - 3}$

$\int \frac{3 \left(4 {y}^{2} - 7 y - 12\right) \mathrm{dy}}{y \left(y + 2\right) \left(y - 3\right)} = 6 \int \frac{\mathrm{dy}}{y} + \frac{27}{5} \int \frac{\mathrm{dy}}{y + 2} + \frac{3}{5} \int \frac{\mathrm{dy}}{y - 3}$

=6ln(∣y∣)+27/5ln(∣y+2∣)+3/5ln(∣y-3∣)+C