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

1 Answer
Jul 18, 2016

#\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: enter image source here

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#.