# `int (x^3+3x-4)/(x^3-4x^2+4x) dx` Use partial fractions to find the indefinite integral

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`