Question

How do you integrate `(5x^2-9x)/((x-4)(x-1)^2)` using partial fractions?

Answer 1

`int (5x^2-9x)/((x-4)(x-1)^2) dx = 44/9 ln abs(x-4) - 4/(3(x-1)) + 1/9 ln abs(x-1) + C`

Explanation:

`(5x^2-9x)/((x-4)(x-1)^2) = A/(x-4)+B/(x-1)^2+C/(x-1)`

`color(white)((5x^2-9x)/((x-4)(x-1)^2)) = (A(x-1)^2+B(x-4)+C(x-4)(x-1))/((x-4)(x-1)^2)`

`color(white)((5x^2-9x)/((x-4)(x-1)^2)) = (A(x^2-2x+1)+B(x-4)+C(x^2-5x+4))/((x-4)(x-1)^2)`

`color(white)((5x^2-9x)/((x-4)(x-1)^2)) = ((A+C)x^2+(-2A+B-5C)x+(A-4B+4C))/((x-4)(x-1)^2)`

Equating coefficients we get this system of linear equations:

`{ (A+C=5), (-2A+B-5C=-9), (A-4B+4C=0) :}`

Adding all three equations, we find:

`-3B = -4`

So `color(blue)(B=4/3)`

Adding twice the first equation to the second, we get:

`B-3C=1`

Hence:

`3C = B-1 = 4/3-1 = 1/3`

So `color(blue)(C=1/9)`

Then from the first equation:

`color(blue)(A=)5-C = 5-1/9 = color(blue)(44/9)`

So:

`int (5x^2-9x)/((x-4)(x-1)^2) dx = int (44/9 * 1/(x-4)+4/3 * 1/(x-1)^2 +1/9 * 1/(x-1)) dx`

`color(white)(int (5x^2-9x)/((x-4)(x-1)^2) dx) = 44/9 ln abs(x-4) - 4/(3(x-1)) + 1/9 ln abs(x-1) +`C