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

Aug 23, 2017

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

$= - \frac{1}{5} \ln \left\mid x + 4 \right\mid - \frac{3}{10} \ln \left\mid x - 1 \right\mid + \frac{1}{2} \ln \left\mid x - 3 \right\mid + C$

#### Explanation:

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

We can find $A$, $B$ and $C$ using Oliver Heaviside's cover up method:

$A = \frac{2 \left(\textcolor{b l u e}{- 4}\right) + 1}{\left(\left(\textcolor{b l u e}{- 4}\right) - 1\right) \left(\left(\textcolor{b l u e}{- 4}\right) - 3\right)} = \frac{- 7}{\left(- 5\right) \left(- 7\right)} = - \frac{1}{5}$

$B = \frac{2 \left(\textcolor{b l u e}{1}\right) + 1}{\left(\left(\textcolor{b l u e}{1}\right) + 4\right) \left(\left(\textcolor{b l u e}{1}\right) - 3\right)} = \frac{3}{\left(5\right) \left(- 2\right)} = - \frac{3}{10}$

$C = \frac{2 \left(\textcolor{b l u e}{3}\right) + 1}{\left(\left(\textcolor{b l u e}{3}\right) + 4\right) \left(\left(\textcolor{b l u e}{3}\right) - 1\right)} = \frac{7}{\left(7\right) \left(2\right)} = \frac{1}{2}$

So:

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

$= \int - \frac{1}{5} \cdot \frac{1}{x + 4} - \frac{3}{10} \cdot \frac{1}{x - 1} + \frac{1}{2} \cdot \frac{1}{x - 3} \mathrm{dx}$

$= - \frac{1}{5} \ln \left\mid x + 4 \right\mid - \frac{3}{10} \ln \left\mid x - 1 \right\mid + \frac{1}{2} \ln \left\mid x - 3 \right\mid + C$