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

Question

1217 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-22T12:17:13

The answer is #=1/3ln(|x+1|)-1/3ln(|x-2|)-2/(x-2)+C#

Perform the decomposition into partial fractions

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

#=(A(x-2)^2+B(x+1)(x-2)+C(x+1))/(((x+1)(x-2)^2))#

The denominators are the same, compare the numerators

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

Let #x=-1#, #=>#, #3=9A#, #=>#, #A=1/3#

Let #x=2#, #=>#, #6=3C#, #=>#, #C=2#

Coefficients of #x^2#

#0=A+B#, #=>#, #B=-1/3#

Therefore,

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

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

#=1/3ln(|x+1|)-1/3ln(|x-2|)-2/(x-2)+C#