How do you use partial fraction decomposition to decompose the fraction to integrate #(x^2-x-6)/(x^3 +3x)#?

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Answer 1 · 2015-08-22T01:34:34

#(x^2-x-6)/(x^3 +3x) = (-2)/x + (3x-1)/(x^2+3)#

#(x^2-x-6)/(x^3 +3x)#

First we need to factor the denominator into quadratic and linear polynomials that are irreducible using Real coefficients.

#x^3+3x = x(x^2+3)#
Neither #x# nor #x^2 + 3# can be further factored with Real coefficients.

So we need to find #A, B, "and " C# to make:

#A/x +(Bx+C)/(x^2+3) = (x^2-x-6)/(x^3 +3x)#

Clear the denominators or get a common denominator on the left to see that we need:

#Ax^2 +3A +Bx^2+Cx = x^2 - x -6#

#(A+B)x^2 +(C)x+(3A) = 1x^2 -1x-6#

Setting the coefficients equal to each other, we need to solve:

#A+B=1#
#C = -1#
#3A = -6#

We can quickly see that we need #A = -2# and #C = -1#, so the first equation becomes:
#-2+B = 1#, so #B =3#

#A/x +(Bx+C)/(x^2+3) = (-2)/x +(3x-1)/(x^2+3) #