How do you evaluate the integral #int (2x-3)/((x-1)(x+4))#?

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Answer 1 · 2018-03-08T22:48:46

#int (2x-3)/((x-1)(x+4)) dx = -1/5 ln abs (x-1)+11/5 ln abs (x+4) +C#

Use partial fractions decomposition:

#(2x-3)/((x-1)(x+4)) = A/(x-1)+B/(x+4)#

#2x-3 = A(x+4)+B(x-1)#

#2x-3 = (A+B)x +4A-B#

#{(A+B=2),(4A-B=-3):}#

#{(5A=-1),(B=-A+2):}#

#{(A=-1/5),(B=11/5):}#

#(2x-3)/((x-1)(x+4)) = -1/(5(x-1))+11/(5(x+4))#

#int (2x-3)/((x-1)(x+4)) dx = -1/5 int dx/(x-1)+11/5 int dx/(x+4)#

#int (2x-3)/((x-1)(x+4)) dx = -1/5 ln abs (x-1)+11/5 ln abs (x+4) +C#