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

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Answer 1 · 2018-03-22T01:54:00

The integral equals #3ln|x + 2| -2 ln|x + 1| + C#

Note that the factoring of #x^2 + 3x + 2# is #(x + 2)(x + 1)#. Therefore:

#A/(x +2)+ B/(x + 1) = (x -1)/((x +2)(x + 1))#

#A(x + 1) + B(x + 2) = x - 1#

#Ax + A + Bx+ 2B = x - 1#

#(A+ B)x + (A + 2B) = x- 1#

We now have a system of equations: #{(A+ B = 1), (A + 2B = -1):}#

We can readily solve through elimination (subtract the second equation from the first to get the following):

#-B = 2#

#B = -2#

It is now clear that #A - 2 = 1 -> A = 3#. The integral becomes

#I = int 3/(x+ 2) - 2/(x + 1)dx#

We can now easily integrate.

#I = 3ln|x + 2| -2 ln|x + 1| + C#

Hopefully this helps!