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

1 Answer
Jul 7, 2016

$= \ln \left(x + 1\right) - \ln \left(x + 2\right) + C$

Explanation:

$\frac{\textcolor{b l u e}{1}}{\left(x + 1\right) \left(x + 2\right)} = \frac{A}{x + 1} + \frac{B}{x + 2}$

$= \frac{\textcolor{b l u e}{A \left(x + 2\right) + B \left(x + 1\right)}}{\left(x + 1\right) \left(x + 2\right)}$

let x = -2, B(-2+1) = 1, B = -1
let x = -1, A(-1+2) = 1, A = 1

so the integrand follows

$= \int \mathrm{dx} q \quad \frac{1}{x + 1} - \frac{1}{x + 2}$

$= \ln \left(x + 1\right) - \ln \left(x + 2\right) + C$