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

Jun 16, 2017

The answer is $= - 2 \ln | \left(x + 5\right) | - \ln | \left(x + 3\right) | + 3 \ln | \left(x + 4\right) | + C$

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{x + 1}{\left(x + 5\right) \left(x + 3\right) \left(x + 4\right)} = \frac{A}{x + 5} + \frac{B}{x + 3} + \frac{C}{x + 4}$

$= \frac{A \left(x + 3\right) \left(x + 4\right) + B \left(x + 5\right) \left(x + 4\right) + C \left(x + 5\right) \left(x + 3\right)}{\left(x + 5\right) \left(x + 3\right) \left(x + 4\right)}$

The denominators are the same, we compare the numerators

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

Let $x = - 5$, $\implies$,

$- 4 = - 2 \cdot - 1 \cdot A$, $\implies$, $A = - 2$

Let $x = - 3$, $\implies$,

$- 2 = 2 \cdot 1 \cdot B$, $\implies$, $B = - 1$

Let $x = - 4$, $\implies$,

$- 3 = 1 \cdot - 1 \cdot C$, $\implies$, $C = 3$

So,

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

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

$= - 2 \ln | \left(x + 5\right) | - \ln | \left(x + 3\right) | + 3 \ln | \left(x + 4\right) | + C$