We need
intx^ndx=x^(n+1)/(n+1)+C (x!=-1)
intdx/x=ln∣x∣+C
We start by factorising the denominator
Let f(x)=x^3-6x^2+12x-8
f(2)=8-24+24-8=0
Therefore,
(x-2) is a factor
To find the other factors, we do a long division
color(white)(aaaa)x^3-6x^2+12x-8color(white)(aaaa)∣x-2
color(white)(aaaa)x^3-2x^2color(white)(aaaaaaaaaaaaa)∣x^2-4x+4
color(white)(aaaaa)0-4x^2+12x
color(white)(aaaaaaa)-4x^2+8x
color(white)(aaaaaaaaa)-0+4x-8
color(white)(aaaaaaaaaaaaa)+4x-8
color(white)(aaaaaaaaaaaaaa)+0-0
Therefore,
x^3-6x^2+12x-8=(x-2)(x^2-4x+4)=(x-2)(x-2)(x-2)
so, we can do the decomposition into partial fractions
(x(2x-9))/(x^3-6x^2+12x-8)=(x(2x-9))/(x-2)^3
=A/(x-2)^3+B/(x-2)^2+C/(x-2)
=(A+B(x-2)+C(x-2)^2)/(x-2)^3
x(2x-9)=A+B(x-2)+C(x-2)^2
Let x=2, =>, -10=A
Coefficients of x^2, =>, 2=C
Let x=0, =>, 0=A-2B+4C
=>, 2B=A+4C=-10+8=-2, =>, B=-1
so,
(x(2x-9))/(x^3-6x^2+12x-8)=-10/(x-2)^3-1/(x-2)^2+2/(x-2)
Therefore,
int(x(2x-9)dx)/(x^3-6x^2+12x-8)=-10intdx/(x-2)^3-intdx/(x-2)^2+2intdx/(x-2)
=5/(x-2)^2+1/(x-2)+2ln(∣x-2∣)+ C
