How do you integrate #x / (x² - x - 2)# using partial fractions?
1 Answer
Jun 4, 2017
Explanation:
First, separate the fraction using partial fraction separation:
#x/(x^2-x-2) = x/((x-2)(x+1))#
#x/((x-2)(x+1))=A/(x-2) + B/(x+1)#
#x = A(x+1) + B(x-2)# Set
#x=2# to solve for A:
#2 = A(2+1) + B(2-2)#
#2 = 3A#
#2/3 = A# Set
#x=-1# to solve for B:
#-1 = A(-1+1) + B(-1-2)#
#-1 = -3B#
#1/3 = B#
Therefore, our fraction separates into:
#x/(x^2-x+2) = 2/(3(x-2)) + 1/(3(x+1))#
Finally, integrate:
#int(2/(3(x-2))+1/(3(x+1)))dx#
#= 2/3ln|x-2| + 1/3ln|x+1| + C#
#= 1/3(2ln|x-2| + ln|x+1|) + C#
#= 1/3ln|(x-2)^2(x+1)| + C#
Final Answer
