# How do you integrate (x^(3)+5) / (x^(2)+5x+6) dx using partial fractions?

Jul 18, 2016

$\setminus \int \setminus \frac{{x}^{3} + 5}{{x}^{2} + 5 x + 6} \mathrm{dx} = \setminus \frac{1}{2} \left(x - 10\right) x - 3 \ln \left(x + 2\right) + 22 \ln \left(x + 3\right) + c$

#### Explanation:

Firstly, note that the degree on the numerator is higher than the degree on the denominator.

Thus, we can use polynomial long division to get it into a form where the degree on the numerator is lower than the denominator. From that, we can use partial fractions to simplify.

Here's how:

So, now, we consider the fractional bit, namely, $\setminus \frac{19 x + 35}{{x}^{2} + 5 x + 6}$.

$\setminus \frac{19 x + 35}{{x}^{2} + 5 x + 6} \equiv \setminus \frac{A}{x + 2} + \setminus \frac{B}{x + 3}$.

i.e., $19 x + 35 \equiv A \left(x + 3\right) + B \left(x + 2\right)$.

And by the comparison of coefficients on both sides, we have

$19 = A + B$ and $35 = 3 A + 2 B$

Which yields the solution $A = - 3$ and $B = 22$.

Thus, we can finally write the integrand as...

$\setminus \frac{{x}^{3} + 5}{{x}^{2} + 5 x + 6} \equiv x - 5 - \setminus \frac{3}{x + 2} + \setminus \frac{22}{x + 3}$

Integrating this...

$\int \setminus \frac{{x}^{3} + 5}{{x}^{2} + 5 x + 6} \mathrm{dx} = \int x - 5 - \setminus \frac{3}{x + 2} + \setminus \frac{22}{x + 3} \mathrm{dx}$

i.e., $= \setminus \frac{{x}^{2}}{2} - 5 x - 3 \ln \left(x + 2\right) + 22 \ln \left(x + 3\right) + c$
$= \setminus \frac{1}{2} \left(x - 10\right) x - 3 \ln \left(x + 2\right) + 22 \ln \left(x + 3\right) + c$.