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

Jul 11, 2016

$- \frac{1}{20} \ln | x - 1 | + \frac{13}{4} \ln | x + 3 | - \frac{16}{5} \ln | x + 4 | + C .$

#### Explanation:

Put $I = \int \frac{3 x - 4}{\left(x - 1\right) \left(x + 3\right) \left(x + 4\right)} \mathrm{dx} . ,$ and, notice that the $D r .$ of the rational integrand is made of non-repeated linear factors. In such case, we have to find, $A , B , D \in \mathbb{R}$ such that,
$\frac{3 x - 4}{\left(x - 1\right) \left(x + 3\right) \left(x + 4\right)} = \frac{A}{x - 1} + \frac{B}{x + 3} + \frac{D}{x + 4} \ldots \ldots \ldots \ldots \left(i\right)$
$= \frac{A \left(x + 3\right) \left(x + 4\right) + B \left(x - 1\right) \left(x + 4\right) + D \left(x - 1\right) \left(x + 3\right)}{\left(x - 1\right) \left(x + 3\right) \left(x + 4\right)} .$

As the $D r s .$ of both sides are same, so must be the $N r s .$ Hence,

$A \left(x + 3\right) \left(x + 4\right) + B \left(x - 1\right) \left(x + 4\right) + D \left(x - 1\right) \left(x + 3\right) = 3 x - 4. \ldots \ldots \ldots . \left(i i\right)$

Although $\left(i i\right)$ holds good $\forall x \in \mathbb{R}$, but, we chose $x = 1 , x = - 3 , \mathmr{and} , x = - 4$ [the reason behind this will soon be clear!] to get,
$A \left(4\right) \left(5\right) + B \cdot 0 + D \cdot 0 = 3 - 4 \Rightarrow A = - \frac{1}{20.}$
$A \cdot 0 + B \left(- 4\right) \left(1\right) + D \cdot 0 = - 9 - 4 \Rightarrow B = \frac{13}{4.}$
$A \cdot 0 + B \cdot + D \left(- 5\right) \left(- 1\right) = - 12 - 4 \Rightarrow D = - \frac{16}{5.}$ Therefore, by $\left(i\right)$
$I = \int \frac{- \frac{1}{20}}{x - 1} \mathrm{dx} + \int \frac{\frac{13}{4}}{x + 3} \mathrm{dx} + \int \frac{- \frac{16}{5}}{x + 4} \mathrm{dx}$
$I = - \frac{1}{20} \ln | x - 1 | + \frac{13}{4} \ln | x + 3 | - \frac{16}{5} \ln | x + 4 | + C .$

Enjoy Maths.!