Question

How do you integrate `(x^4+x^3+x^2+1)/(x^2+x-2)` using partial fractions?

Answer 1

Because the order of the numerator is greater than the denominator some reduction must be done before expanding the partial fractions. Please see the explanation.

Explanation:

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

Begin reducing by adding 0 to the numerator in the form of `-2x^2 + 2x^2`

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

Break into two fractions:

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

Factor `x^2` from the numerator of the first term:

`(x^2(cancel(x^2 + x -2)))/cancel(x^2 + x - 2) + (2x^2 + x^2 + 1)/(x^2 + x - 2)`

The first term becomes `x^2` and we combine like terms in the numerator of the second fraction:

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

Add 0 to the numerator of the second term in the form of `3x - 6 - 3x + 6`:

`x^2 + (3x^2 + 3x - 6 - 3x + 6 + 1)/(x^2 + x - 2)`

Break the second term into two fractions:

`x^2 + (3x^2 + 3x - 6)/(x^2 + x - 2) + (-3x + 6 + 1)/(x^2 + x - 2)`

Remove a factor of 3 from the first numerator and combine like terms in the last fraction:

`x^2 + (3(cancel(x^2 + x - 2)))/cancel(x^2 + x - 2) + (7 -3x)/(x^2 + x - 2)`

The second term becomes 3:

`x^2 + 3 + (7 -3x)/(x^2 + x - 2)`

Partial Fraction Expansion of the last term:

`(7 -3x)/(x^2 + x - 2) = A/(x + 2) + B/(x - 1)`

`7 - 3x = A(x - 1) + B(x + 2)`

Make B disappear by letting x = -2

`7 - 3(-2) = A(-2 - 1) + B(-2 + 2)`

`A` = -13/3

Make A disappear by Letting x = 1:

`7 - 3(1) = A(1 - 1) + B(1 + 2)`

`B = 4/3`

Check

`(-13/3)/(x + 2) + (4/3)/(x - 1)`

`(-13/3)/(x + 2)(x - 1)/(x - 1) + (4/3)/(x - 1)(x + 2)/(x + 2)`

`((-13/3)(x - 1)+ (4/3)(x + 2))/((x + 2)(x - 1))`

`(7 - 3x)/((x + 2)(x - 1))`

This checks

Returning to the main problem:

`int(x^4 + x^3 + x^2 + 1)/(x^2 + x - 2)dx = intx^2dx + 3intdx - 13/3int1/(x + 2)dx + 4/3int1/(x - 1)dx`

`int(x^4 + x^3 + x^2 + 1)/(x^2 + x - 2)dx = 1/3x^3 + 3x - 13/3ln(x + 2) + 4/3ln(x - 1) + C`