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

Aug 16, 2017

$\int \setminus {x}^{3} / \left({x}^{2} + 4 x + 3\right) \setminus \mathrm{dx} = \frac{1}{2} {x}^{2} - 4 x + \frac{27}{2} \ln | x + 3 | - \frac{1}{2} \ln | x + 1 | + c$

#### Explanation:

We seek:

$I = \int \setminus {x}^{3} / \left({x}^{2} + 4 x + 3\right) \setminus \mathrm{dx}$

Before we consider a partial fraction decomposition we note that the order of the numerator is one degree higher than the order of the denominator (the equivalent of a "top-heavy" fraction). So we must use algebraic long division in order to reduce the order.

The algebraic division is as follows

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus x \setminus \setminus - 4$
${x}^{2} + 4 x + 3 \setminus \overline{\text{)} {x}^{3} \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus} \setminus -$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus {x}^{3} + 4 {x}^{2} + 3 x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \overline{\setminus \setminus 0 - 4 {x}^{2} - 3 x \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus} \setminus \setminus -$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus - 4 {x}^{2} - 16 x - 12$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \overline{\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 0 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 13 x + 12 \setminus \setminus}$

And so we can write the integral as follows:

$I = \int \setminus \left(x - 4\right) + \frac{13 x + 12}{{x}^{2} + 4 x + 3} \setminus \mathrm{dx}$
$\setminus \setminus = \int \setminus \left(x - 4\right) \setminus \mathrm{dx} + \int \setminus \frac{13 x + 12}{\left(x + 3\right) \left(x + 1\right)} \setminus \mathrm{dx}$

We can readily handle the first integral, so now we must deal with the second integral by decomposing the integrand into partial fractions, which will take the form:

$\frac{13 x + 12}{\left(x + 3\right) \left(x + 4\right)} \equiv \frac{A}{x + 3} + \frac{B}{x + 1}$
$\text{ } = \frac{A \left(x + 1\right) + B \left(x + 3\right)}{\left(x + 3\right) \left(x + 1\right)}$

$13 x + 12 = A \left(x + 1\right) + B \left(x + 3\right)$

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

Put $x = - 3 \implies - 27 = - 2 A \implies A = \frac{27}{2}$
Put $x = - 1 \implies - 1 = 2 B \setminus \setminus \setminus \setminus \setminus \implies B = - \frac{1}{2}$

So using partial fraction decomposition we have:

$I = \int \setminus \left(x - 4\right) \setminus \mathrm{dx} + \setminus \int \setminus \frac{\frac{27}{2}}{x + 3} + \frac{- \frac{1}{2}}{x + 1} \setminus \mathrm{dx}$

And now all integrands are readily integratable, so:

$I = \frac{1}{2} {x}^{2} - 4 x + \frac{27}{2} \ln | x + 3 | - \frac{1}{2} \ln | x + 1 | + c$