We need
∫xndx=xn+1n+1+C(x≠−1)
∫dxx=ln∣x∣+C
We start by factorising the denominator
Let f(x)=x3−6x2+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
aaaax3−6x2+12x−8aaaa∣x−2
aaaax3−2x2aaaaaaaaaaaaa∣x2−4x+4
aaaaa0−4x2+12x
aaaaaaa−4x2+8x
aaaaaaaaa−0+4x−8
aaaaaaaaaaaaa+4x−8
aaaaaaaaaaaaaa+0−0
Therefore,
x3−6x2+12x−8=(x−2)(x2−4x+4)=(x−2)(x−2)(x−2)
so, we can do the decomposition into partial fractions
x(2x−9)x3−6x2+12x−8=x(2x−9)(x−2)3
=A(x−2)3+B(x−2)2+Cx−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 x2, ⇒, 2=C
Let x=0, ⇒, 0=A−2B+4C
⇒, 2B=A+4C=−10+8=−2, ⇒, B=−1
so,
x(2x−9)x3−6x2+12x−8=−10(x−2)3−1(x−2)2+2x−2
Therefore,
∫x(2x−9)dxx3−6x2+12x−8=−10∫dx(x−2)3−∫dx(x−2)2+2∫dxx−2
=5(x−2)2+1x−2+2ln(∣x−2∣)+C
