(4x^2-7x-12)/((x)(x+2)(x-3))=A/x+B/(x+2)+C/(x-3)=4x2−7x−12(x)(x+2)(x−3)=Ax+Bx+2+Cx−3=
=(A(x+2)(x-3)+B(x)(x-3)+C(x)(x+2))/((x)(x+2)(x-3))=A(x+2)(x−3)+B(x)(x−3)+C(x)(x+2)(x)(x+2)(x−3)
then:
A(x^2-x-6)+B(x^2-3x)+C(x^2+2x)=4x^2-7x-12A(x2−x−6)+B(x2−3x)+C(x2+2x)=4x2−7x−12
cons. x=0 x=0 then A(-6)+B(0)+C(0)=-12 rArr A=2A(−6)+B(0)+C(0)=−12⇒A=2
x^2-x-6=0rArrx_1=3and x_2=-2x2−x−6=0⇒x1=3andx2=−2
cons. x=-2 x=−2 then A(0)+B(10)+C(0)=18 rArr B=9/5A(0)+B(10)+C(0)=18⇒B=95
cons. x=3 x=3 then A(0)+B(0)+C(15)=3 rArr C=1/5A(0)+B(0)+C(15)=3⇒C=15
finally get to:
(4x^2-7x-12)/((x)(x+2)(x-3))=2/x+9/(5(x+2))+1/(5(x-3))4x2−7x−12(x)(x+2)(x−3)=2x+95(x+2)+15(x−3)
then:
int(4x^2-7x-12)/((x)(x+2)(x-3))dx=int(2/x+9/(5(x+2))+1/(5(x-3)))dx=int2/xdx+int9/(5(x+2))dx+int1/(5(x-3))dx=∫4x2−7x−12(x)(x+2)(x−3)dx=∫(2x+95(x+2)+15(x−3))dx=∫2xdx+∫95(x+2)dx+∫15(x−3)dx=
2ln absx +9/5lnabs(x+2)+1/5lnabs(x-3)+ C2ln|x|+95ln|x+2|+15ln|x−3|+C
