How do you integrate #int (2x+1)/((x+4)(x-1)(x-3)) # using partial fractions?

1 Answer
George C.
Aug 23, 2017

#int (2x+1)/((x+4)(x-1)(x-3)) dx#

#= -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#, #B# and #C# using Oliver Heaviside's cover up method:

#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#

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