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

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Answer 1 · 2016-07-07T14:09:24

$= ln (x+1) - ln (x+2) + C$

$color{blue}{1}/ ((x + 1)(x + 2)) = A/(x+1) + B/(x+2)$

$= (color{blue}{A(x+2) + B (x+1)})/((x+1)(x+2))$

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

so the integrand follows

$= int dx qquad 1/(x+1) - 1/(x+2)$

$= ln (x+1) - ln (x+2) + C$

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