Calculus of a Single Variable, Chapter 8, 8.5, Section 8.5, Problem 15

For the given integral problem: int (x^3+3x-4)/(x^3-4x^2+4x)dx , we may simplify by applying long division since the highest degree of x is the same from numerator and denominator side.
(x^3+3x-4)/(x^3-4x^2+4x) = 1+(4x^2-x-4)/(x^3-4x^2+4x) .
Apply partial fraction decomposition on the expression (4x^2-x-4)/(x^3-4x^2+4x) .
The pattern on setting up partial fractions will depend on the factors of the denominator. For the given problem, the factored form of the denominator will be:
(x^3-4x^2+4x) =(x)(x^2-4x+4)
=(x) (x-2)(x-2) or x(x-2)^2
For the linear factor (x) , we will have partial fraction: A/x
For the repeated linear factor (x-2)^2 , we will have partial fractions: B/(x-2) + C/(x-2)^2 .
The rational expression becomes:
(4x^2-x-4)/(x^3-4x^2+4x) =A/x +B/(x-2) + C/(x-2)^2
Multiply both side by the LCD =x(x-2)^2 :
((4x^2-x-4)/(x^3-4x^2+4x)) (x(x-2)^2)=(A/x +B/(x-2) + C/(x-2)^2)(x(x-2)^2)
4x^2-x-4=A*(x-2)^2+B*(x(x-2)) + C*x
We apply zero-factor property on x(x-2)^2 to solve for value we can assign on x.
x=0
x-2 = 0 then x=2 .
To solve for A , we plug-in x=0 :
4*0^2-0-4=A*(0-2)^2+B*(0(0-2)) + C*0
0-0-4 = A*(-2)^2 +0 +0
-4 =4A
-4/4 =(4A)/4
A =-1
To solve for C , we plug-in x=2 :
4*2^2-2-4=A*(2-2)^2+B*(2(2-2)) + C*2
16-2-4 = A*0 +B*0 +2C
10= 0 + 0 +2C
10 =2C
(10)/2= (2C)/2
C=5
To solve for B, plug-in x=1 ,A=-1 , and C=5 :

4*1^2-1-4=(-1)*(1-2)^2+B*(1(1-2)) + 5*1
4-1-4= (-1)*(-1)^2+B(1*(-1)) +5
-1= -1-B +5
-1= -B+4
-1-4= -B
-5=-B
(-5)/(-1) = (-B)/(-1)
B =5
Plug-in A = -1 , B =5, and C=5 , we get the partial decomposition:
(4x^2-x-4)/(x^3-4x^2+4x) =-1/x +5/(x-2) + 5/(x-2)^2
Then the integrand becomes:
(x^3+3x-4)/(x^3-4x^2+4x) = 1+(4x^2-x-4)/(x^3-4x^2+4x) .
=1-1/x +5/(x-2) + 5/(x-2)^2
Apply the basic integration property:int (u+-v) dx = int (u) dx +- int (v) dx .
int (x^3+3x-4)/(x^3-4x^2+4x) dx = int [1-1/x +5/(x-2) + 5/(x-2)^2] dx
=int1 dx - int 1/x dx +int 5/(x-2)dx + int 5/(x-2)^2dx

Apply basic integration property: int(a) dx = ax+C
int1 dx = 1x or x
Apply integration formula for logarithm: int 1/u du = ln|u|+C .
int 1/x dx=ln|x|
int 5/(x-2)dx= int 5/udu
= 5ln|u|
=5 ln|x-2|
Note: Let u =x-2 then du = dx .
Apply the Power Rule for integration: int (u^n) dx =u^(n+1)/ (n+1) +C .
int 5/(x-2)^2dx=int 5/u^2du
=int 5u^(-2)du
= 5 * u^(-2+1)/(-2+1)
= 5* u^-1/(-1)
= -5/u
= -5/(x-2)
Note: Let u =x-2 then du = dx
Combining the results, we get the indefinite integral as:
int (x^3+3x-4)/(x^3-4x^2+4x)dx =x-ln|x| +5 ln|x-2|-5/(x-2)+C