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

Dec 4, 2017

$\int \frac{\left(5 x + 1\right) \cdot \mathrm{dx}}{\left(x + 3\right) \left(x + 2\right) \left(x - 4\right)}$

=$\frac{1}{2} \cdot \ln \left(x - 4\right) + \frac{3}{2} \cdot \ln \left(x + 2\right) - 2 \ln \left(x + 3\right) + C$

#### Explanation:

I decomposed integrand into basic fractions,

$\frac{5 x + 1}{\left(x + 3\right) \left(x + 2\right) \left(x - 4\right)}$

=$\frac{A}{x + 3} + \frac{B}{x + 2} + \frac{C}{x - 4}$

After expanding denominator,

$A \left(x + 2\right) \left(x - 4\right) + B \left(x + 3\right) \left(x - 4\right) + C \left(x + 3\right) \left(x + 2\right) = 5 x + 1$

After setting $x = - 3$, $7 A = - 14$, so $A = - 2$

After setting $x = - 2$, $- 6 B = - 9$, so $B = \frac{3}{2}$

After setting $x = 4$, $42 C = 21$, so $C = \frac{1}{2}$

Thus,

$\int \frac{\left(5 x + 1\right) \cdot \mathrm{dx}}{\left(x + 3\right) \left(x + 2\right) \left(x - 4\right)}$

=$- 2 \cdot \int \frac{\mathrm{dx}}{x + 3} + \frac{3}{2} \cdot \int \frac{\mathrm{dx}}{x + 2} + \frac{1}{2} \cdot \int \frac{\mathrm{dx}}{x - 4}$

=$\frac{1}{2} \cdot \ln \left(x - 4\right) + \frac{3}{2} \cdot \ln \left(x + 2\right) - 2 \ln \left(x + 3\right) + C$