How do you integrate # x/(x^3-x^2-6x)# using partial fractions?
1 Answer
Explanation:
Quick version
Note that both the numerator and denominator are divisible by
#x/(x^3-x^2-6x) = 1/(x^2-x-6) = 1/((x-3)(x+2)) = 1/(5(x-3))-1/(5(x+2))#
So:
#int x/(x^3-x^2-6x) dx = int 1/(5(x-3))-1/(5(x+2)) dx#
#color(white)(int x/(x^3-x^2-6x) dx) = 1/5 ln abs(x-3)-1/5 ln abs(x+2) + C#
Notes
#1/((x-3)(x+2)) = A/(x-3) + B/(x+2)#
#color(white)(1/((x-3)(x+2))) = (A(x+2)+B(x-3))/((x-3)(x+2))#
#color(white)(1/((x-3)(x+2))) = ((A+B)x+(2A-3B))/((x-3)(x+2))#
Equating coefficients, we have:
#{ (A+B = 0), (2A-3B = 1) :}#
Adding
#5A = 1#
Hence
So:
#1/((x-3)(x+2)) = 1/(5(x-3)) - 1/(5(x+2))#
Alternatively, we could find
#A = 1/(color(blue)(3)+2) = 1/5#
#B = 1/(color(blue)(-2)-3) = 1/(-5) = -1/5#
The method I actually used was:
If
then when we express in terms of a common denominator we will need
If try
Hence