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-8##color(white)(aaaa)##∣##x-2#
#color(white)(aaaa)##x^3-2x^2##color(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#
