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

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Answer 1 · 2016-12-15T03:31:43

$int dx/(x^2+3x+2)= ln |(x+1)/(x+2)|+C$

First you need to factorize the denominator:

$x^2+3x+2 = (x+1)(x+2)$

Now put:

$frac 1 ((x+1)(x+2)) = A/(x+1)+B/(x+2)$

$frac 1 ((x+1)(x+2)) = frac (A(x+2)+B(x+1)) ((x+1)(x+2)) = frac (Ax+2A+Bx+B) ((x+1)(x+2))$

As the denominators are the same, so must be the numerators:

$(Ax+2A+Bx+B) = 1$

and equating the coefficients of the same degree in $x$ :

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

Solving this system we get:

$A=1$ , $B=-1$

so that:

$int dx/(x^2+3x+2)= int (1/(x+1)-1/(x+2))dx= int dx/(x+1) - int dx/(x+2)= ln|x+1|-ln|x+2| +C = ln |(x+1)/(x+2)|+C$

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