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

Mar 27, 2017

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

#### Explanation:

Write the rational function as a sum of partial fractions:

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

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

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

As the denominators are equal we can equate the numerators:

$1 = \left(A + B\right) x + 2 A + B$

and equating the coefficients with the same degree in $x$ we have:

$\left\{\begin{matrix}A + B = 0 \\ 2 A + B = 1\end{matrix}\right.$

$\left\{\begin{matrix}B = - A \\ 2 A - A = 1\end{matrix}\right.$

$\left\{\begin{matrix}A = 1 \\ B = - 1\end{matrix}\right.$

So that:

$\frac{1}{\left(x + 1\right) \left(x + 2\right)} = \frac{1}{x + 1} - \frac{1}{x + 2}$

and integrating:

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

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

and using the properties of logarithms:

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