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

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Answer 1 · 2017-02-16T11:20:17

The answer is #=7/18ln(|x+4|)+11/18ln(|x-2|)-1/(3(x-2))+C#

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let #x=-4#, #=>#, #14=36A#, #=>#, #A=7/18#

Let #x=2#, #=>#, #2=6B#, #=>#, #B=1/3#

Coefficients of #x^2#,

#1=A+C#, #=>#, #C=1-A=1-7/18=11/18#

Therefore,

#(x^2-2)/((x+4)(x-2)^2)=(7/18)/(x+4)+(1/3)/(x-2)^2+(11/18)/(x-2)#

#int((x^2-2)dx)/((x+4)(x-2)^2)=7/18intdx/(x+4)+1/3intdx/(x-2)^2+11/18intdx/(x-2)#

#=7/18ln(|x+4|)+11/18ln(|x-2|)-1/(3(x-2))+C#