Let's work out the decomposition into partial fractions
(x-5x^2)/((x+2)(x-1)(x-3))x−5x2(x+2)(x−1)(x−3)
=A/(x+2)+B/(x-1)+C/(x-3)=Ax+2+Bx−1+Cx−3
=(A(x-1)(x-3)+B(x+2)(x-3)+C(x+2)(x-1))/((x+2)(x-1)(x-3))=A(x−1)(x−3)+B(x+2)(x−3)+C(x+2)(x−1)(x+2)(x−1)(x−3)
The denominators are the same, we can compare the numerators
x-5x^2=A(x-1)(x-3)+B(x+2)(x-3)+C(x+2)(x-1)x−5x2=A(x−1)(x−3)+B(x+2)(x−3)+C(x+2)(x−1)
Let x=-2x=−2, =>⇒, -22=15A−22=15A, =>⇒, A=-22/15A=−2215
Let x=1x=1, =>⇒, -4=-6B−4=−6B, =>⇒, B=2/3B=23
Let x=3x=3, =>⇒, -42=10C−42=10C, =>⇒, C=-21/5C=−215
Therefore,
(x-5x^2)/((x+2)(x-1)(x-3))=(-22/15)/(x+2)+(2/3)/(x-1)+(-21/5)/(x-3)x−5x2(x+2)(x−1)(x−3)=−2215x+2+23x−1+−215x−3
So,
int((x-5x^2)dx)/((x+2)(x-1)(x-3))=-22/15intdx/(x+2)+2/3intdx/(x-1)-21/5intdx/(x-3)∫(x−5x2)dx(x+2)(x−1)(x−3)=−2215∫dxx+2+23∫dxx−1−215∫dxx−3
=-22/15ln(|x+2|)+2/3ln(|x-1|)-21/5ln(|x-3|) + C=−2215ln(|x+2|)+23ln(|x−1|)−215ln(|x−3|)+C
