How do you integrate #x^2/((x-3)(x+2)^2)#?

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Answer 1 · 2017-02-13T15:51:57

The answer is #=9/25ln(|x-3|)+16/25ln(|x+2|)+4/(5(x+2))+C#

Let's perform the decomposition into partial fractions

#x^2/((x-3)(x+2)^2)=A/(x-3)+B/(x+2)^2+C/(x+2)#

#=(A(x+2)^2+B(x-3)+C(x+2)(x-3))/((x-3)(x+2)^2)#

The denominators are the same, we compare the numerators

#x^2=A(x+2)^2+B(x-3)+C(x+2)(x-3)#

Let #x=3#, #=>#, #9=25A#, #=>#, #A=9/25#

Let #x=-2#, #=>#, #4=-5B#, #=>#, ##B=-4/5

Coefficients of #x^2#,

#1=A+C#

#C=1-A=1-9/25=16/25#

Therefore,

#x^2/((x-3)(x+2)^2)=(9/25)/(x-3)+(-4/5)/(x+2)^2+(16/25)/(x+2)#

So,

#int(x^2dx)/((x-3)(x+2)^2)=9/25intdx/(x-3)-4/5intdx/(x+2)^2+16/25intdx/(x+2)#

#=9/25ln(|x-3|)+16/25ln(|x+2|)+4/(5(x+2))+C#