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

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Answer 1 · 2016-11-30T13:24:55

$int frac 1 (x^2+3x+2) dx= ln (frac (x+1) (x+2))$

Factorize the denominator:

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

Now develop in partial fractions using parametric numerators:

$frac 1 (x^2+3x+2) = A/(x+1)+B/(x+2)$

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

$frac 1 (x^2+3x+2) = frac (A(x+2)+B(x+1)) ((x+1)(x+2)) = frac (Ax+2A+Bx+B) ((x+1)(x+2))= frac ((A+B)x+(2A+B)) ((x+1)(x+2))$

So:

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

Solving the system:

$A=1$
$B=-1$

Finally:

$frac 1 (x^2+3x+2) = 1/(x+1)-1/(x+2)$

We are now ready to integrate:

$int frac 1 (x^2+3x+2) dx= int (1/(x+1)-1/(x+2))dx =$

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

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