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

1 Answer
Mar 7, 2016

#x^2/4+(3x)/4+17/40ln |x+1/2|+11/5ln|x-2|+const.#

Explanation:

First we should get in the numerator a polynomial of a grade inferior than the denominator's

By long division:

#" "x^3+0x^2+2x-1" "#|#" "2x^2-3x-2#
#-x^3+3/2x^2+x" "#|____
_____#" "1/2x+3/4#
#" "3/2x^2+3x-1#
#" "-3/2x^2+9/4x+3/2#
#" "# _______
#" "21/4x+1/2=1/4(21x+2)#

So the expression becomes
#=int(x/2+3/4)dx+1/4int (21x+2)/(2x^2-3x-2)dx#

Let's deal with the last part
Finding the zeros of the denominator
#2x^2-3x-2=0#

#Delta=9+16=25# => #sqrt(Delta)=5#
#x=(3+-5)/4# => #x_1=-1/2#; #x_2=2#

Then we can break the second integrand in this way:
#(21x+2)/(2x^2-3x-2)=A/(x+1/2)+B/(x-2)#
To find #A# and #B# let's make #x=0 and -1#

#-1=2A-B/2#
#-19/3=-2A-B/3#
Summing these 2 expressions we get
#-22/3=-5/6B# => #B=44/5#
#-> A=(-1+B/2)/2=(-1+22/5)/2# => #A=17/10#

So the main expression becomes
#=x^2/4+(3x)/4+(1/4)(17/10int dx/(x+1/2)+44/5int dx/(x-2))#
#=x^2/4+(3x)/4+17/40ln |x+1/2|+11/5ln|x-2|+const.#