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

Nov 30, 2016

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

Explanation:

Factorize the denominator:

${x}^{2} + 3 x + 2 = \left(x + 1\right) \left(x + 2\right)$

Now develop in partial fractions using parametric numerators:

$\frac{1}{{x}^{2} + 3 x + 2} = \frac{A}{x + 1} + \frac{B}{x + 2}$

Expand and equate the coefficient of the same order of the left side and right side numerators to determine $A$ and $B$:

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

So:

$A + B = 0$
$2 A + B = 1$

Solving the system:

$A = 1$
$B = - 1$

Finally:

$\frac{1}{{x}^{2} + 3 x + 2} = \frac{1}{x + 1} - \frac{1}{x + 2}$

We are now ready to integrate:

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

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