Question

How do you integrate `(x^(3)+5) / (x^(2)+5x+6) dx` using partial fractions?

Answer 1

`\int \frac{x^3+5}{x^2+5x+6} dx = \frac{1}{2} (x-10)x - 3ln(x+2) +22ln(x+3) + c`

Explanation:

Firstly, note that the degree on the numerator is higher than the degree on the denominator.

Thus, we can use polynomial long division to get it into a form where the degree on the numerator is lower than the denominator. From that, we can use partial fractions to simplify.

Here's how:

So, now, we consider the fractional bit, namely, `\frac{19x+35}{x^2+5x+6}`.

`\frac{19x+35}{x^2+5x+6} -= \frac{A}{x+2} + \frac{B}{x+3}`.

i.e., `19x+35 -= A(x+3) + B(x+2)`.

And by the comparison of coefficients on both sides, we have

`19 = A + B` and `35 = 3A + 2B`

Which yields the solution `A = -3` and `B = 22`.

Thus, we can finally write the integrand as...

`\frac{x^3+5}{x^2+5x+6} -= x-5-\frac{3}{x+2}+\frac{22}{x+3}`

Integrating this...

`int \frac{x^3+5}{x^2+5x+6} dx = int x-5-\frac{3}{x+2}+\frac{22}{x+3} dx`

i.e., `= \frac{x^2}{2} - 5x -3ln(x+2) + 22ln(x+3) + c`
`=\frac{1}{2}(x-10)x - 3ln(x+2) + 22ln(x+3) + c`.