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

Jun 11, 2017

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

Explanation:

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

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

$A = \frac{\left(\textcolor{b l u e}{- 3}\right) - {\left(\textcolor{b l u e}{- 3}\right)}^{2}}{\left(\left(\textcolor{b l u e}{- 3}\right) - 1\right) \left(\left(\textcolor{b l u e}{- 3}\right) + 4\right)} = \frac{- 12}{\left(- 4\right) \left(1\right)} = 3$

$B = \frac{\left(\textcolor{b l u e}{1}\right) - {\left(\textcolor{b l u e}{1}\right)}^{2}}{\left(\left(\textcolor{b l u e}{1}\right) + 3\right) \left(\left(\textcolor{b l u e}{1}\right) + 4\right)} = \frac{0}{\left(4\right) \left(5\right)} = 0$

$C = \frac{\left(\textcolor{b l u e}{- 4}\right) - {\left(\textcolor{b l u e}{- 4}\right)}^{2}}{\left(\left(\textcolor{b l u e}{- 4}\right) + 3\right) \left(\left(\textcolor{b l u e}{- 4}\right) - 1\right)} = \frac{- 20}{\left(- 1\right) \left(- 5\right)} = - 4$

So:

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

$\textcolor{w h i t e}{\int \frac{x - {x}^{2}}{\left(x + 3\right) \left(x - 1\right) \left(x + 4\right)} \mathrm{dx}} = 3 \ln \left\mid x + 3 \right\mid - 4 \ln \left\mid x + 4 \right\mid + C$