Question

What is `int (x^2-5x+4 ) / (x^3-2x +1 )`?

Answer 1

`int (x^2-5x+4)/(x^3-2x+1) dx`

`= ((5-9sqrt(5))/10)ln(abs(x+1/2-sqrt(5)/2)) + ((5+9sqrt(5))/10)ln(abs(x+1/2+sqrt(5)/2)) + C`

Explanation:

`(x^2-5x+4)/(x^3-2x+1)`

`=(color(red)(cancel(color(black)((x-1))))(x-4))/(color(red)(cancel(color(black)((x-1))))(x^2+x-1)`

`=(x-4)/(x^2+x-1)`

`=(x-4)/((x+1/2)^2-5/4)`

`=(x-4)/((x+1/2-sqrt(5)/2)(x+1/2+sqrt(5)/2))`

`=A/(x+1/2-sqrt(5)/2) + B/(x+1/2+sqrt(5)/2)`

`=(A(x+1/2+sqrt(5)/2)+B(x+1/2-sqrt(5)/2))/(x^2+x-1)`

`=((A+B)x + ((1/2+sqrt(5)/2)A+(1/2-sqrt(5)/2)B))/(x^2+x-1)`

Equating coefficients:

`{ (A+B=1), ((1/2+sqrt(5)/2)A+(1/2-sqrt(5)/2)B = -4) :}`

Subtract `1/2` of the first equation from the second to get:

`sqrt(5)/2(A-B) = -9/2`

Multiply both sides by `2/sqrt(5)` to find:

`A-B = -9/sqrt(5) = -(9sqrt(5))/5`

Add this to the first equation to find:

`2A = 1-(9sqrt(5))/5 = (5-9sqrt(5))/5`

Hence:

`A = (5-9sqrt(5))/10`

`B = (5+9sqrt(5))/10`

So:

`int (x^2-5x+4)/(x^3-2x+1) dx`

`= int ((5-9sqrt(5))/(10(x+1/2-sqrt(5)/2)) + (5+9sqrt(5))/(10(x+1/2+sqrt(5)/2))) dx`

`= ((5-9sqrt(5))/10)ln(abs(x+1/2-sqrt(5)/2)) + ((5+9sqrt(5))/10)ln(abs(x+1/2+sqrt(5)/2)) + C`