How do you integrate int (2x+1)/((x+4)(x-1)(x-3)) using partial fractions?
1 Answer
Aug 23, 2017
= -1/5 ln abs(x+4)- 3/10ln abs(x-1)+1/2ln abs(x-3)+C
Explanation:
(2x+1)/((x+4)(x-1)(x-3)) = A/(x+4)+B/(x-1)+C/(x-3)
We can find
A = (2(color(blue)(-4))+1)/(((color(blue)(-4))-1)((color(blue)(-4))-3)) = (-7)/((-5)(-7)) = -1/5
B = (2(color(blue)(1))+1)/(((color(blue)(1))+4)((color(blue)(1))-3)) = 3/((5)(-2)) = -3/10
C = (2(color(blue)(3))+1)/(((color(blue)(3))+4)((color(blue)(3))-1)) = 7/((7)(2)) = 1/2
So:
int (2x+1)/((x+4)(x-1)(x-3)) dx
= int -1/5*1/(x+4)-3/10*1/(x-1)+1/2*1/(x-3) dx
= -1/5 ln abs(x+4)- 3/10ln abs(x-1)+1/2ln abs(x-3)+C
