Question

How do you use partial fractions to find the integral `int (x^2-x+2)/(x^3-x^2+x-1)dx`?

Answer 1

`ln|x - 1| - arctanx + C`

Explanation:

The denominator can be factored as `x^2(x - 1) + 1(x- 1) = (x^2 + 1)(x- 1)`.

`(Ax + B)/(x^2 + 1) + C/(x- 1) = (x^2 - x + 2)/((x^2 + 1)(x - 1))`

`(Ax + B)(x- 1) + C(x^2 + 1) = x^2 - x + 2`

`Ax^2 + Bx - Ax - B + Cx^2 + C = x^2 - x + 2`

`(A + C)x^2 + (B - A)x + (C - B) = x^2 - x + 2`

Then, we can write a systems of equations.

`{(A + C = 1), (B - A = -1), (C - B = 2):}`

Solve to get `A = 0, B = -1, C= 1`.

Therefore, the partial fraction decomposition is `-1/(x^2 + 1) + 1/(x - 1)`.

The integral becomes `int(1/(x - 1) - 1/(x^2 + 1))dx`.

Note that `d/dx(arctanx) = 1/(x^2 + 1)dx` and that `d/dx(lnx) = 1/xdx`.

Therefore, the integral is `ln|x - 1| - arctanx + C`.

Hopefully this helps!