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

Question

2733 views around the world

You can reuse this answer
Creative Commons License

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))#