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,

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

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

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