# How do you integrate int (x^2 +x -16) / [(x+1)(x-3)^2] dx using partial fractions?

Aug 4, 2017

$\int \setminus \frac{{x}^{2} + x - 16}{\left(x + 1\right) {\left(x - 3\right)}^{2}} \setminus \mathrm{dx} = - \ln | x + 1 | + 2 \ln | x - 3 | + \frac{1}{x - 3} + C$

#### Explanation:

We have:

$I = \int \setminus \frac{{x}^{2} + x - 16}{\left(x + 1\right) {\left(x - 3\right)}^{2}} \setminus \mathrm{dx}$

We can decompose the integrand into partial fraction, which will take the form:

$\frac{{x}^{2} + x - 16}{\left(x + 1\right) {\left(x - 3\right)}^{2}} \equiv \frac{A}{x + 1} + \frac{B}{x - 3} + \frac{C}{x - 3} ^ 2$
$\text{ } = \frac{A {\left(x - 3\right)}^{2} + B \left(x + 1\right) \left(x - 3\right) + C \left(x + 1\right)}{\left(x + 1\right) {\left(x - 3\right)}^{2}}$

${x}^{2} + x - 16 \equiv A {\left(x - 3\right)}^{2} + B \left(x + 1\right) \left(x - 3\right) + C \left(x + 1\right)$

Where $A , B , C$ are constants that are to be determined. We can find them by substitutions (In practice we do this via the "cover up" method:

Put $x = - 1 \implies - 16 = A {\left(- 4\right)}^{2} \implies A = - 1$
Put $x = \setminus \setminus \setminus \setminus 3 \implies - 4 = 4 C \implies C = - 1$
Put $x = \setminus \setminus \setminus \setminus 0 \implies - 16 = 9 A - 3 B + C \implies B = 2$

So using partial fraction decomposition we have:

$I = \int \setminus - \frac{1}{x + 1} + \frac{2}{x - 3} - \frac{1}{x - 3} ^ 2 \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus - \frac{1}{x + 1} \setminus \mathrm{dx} + 2 \setminus \int \setminus \frac{1}{x - 3} \setminus \mathrm{dx} - \int \setminus \frac{1}{x - 3} ^ 2 \setminus \mathrm{dx}$

And now all integrands are readily integratable, so:

$I = - \ln | x + 1 | + 2 \ln | x - 3 | + \frac{1}{x - 3} + C$