How do you integrate # x^2/ [x^2+x+4]# using partial fractions?

1 Answer
Nov 9, 2016

#int frac{x^2}{x^2+x+4} "d"x #

#= x - 1/2 ln(x^2+x+4) + 7/sqrt15 tan^{-1}(frac{2x + 1}{sqrt15}) + c#

where #c# is the constant of integration.

Explanation:

First, write out the partial fractions

#frac{x^2}{x^2+x+4} = frac{x^2+(x+4) - (x+4)}{x^2+x+4}#

#= 1 - frac{x+4}{x^2+x+4}#

Fractions of the form #(f'(x))/f(x)# integrate to become #ln(f(x))#. Therefore, suppose #f(x) = x^2 + x +4#, then

#f'(x) = 2x + 1#

Rewriting the partial fraction,

#frac{x^2}{x^2+x+4} = 1 - 1/2 frac{2(x+4)}{x^2+x+4}#

#= 1 - 1/2 frac{2x+8}{x^2+x+4}#

#= 1 - 1/2 frac{2x+1}{x^2+x+4} + frac{7}{2} frac{1}{x^2+x+4}#

The last bit requires completing the square, as fractions of the form #1/((x-b)^2 +a^2)# integrate to form #1/a tan^{-1}((x-b)/a)#.

#x^2 + x + 4 = (x + 1/2)^2 + 15/4#

Rewriting the partial fraction,

#frac{x^2}{x^2+x+4} = 1 - 1/2 frac{2x+1}{x^2+x+4} + frac{14}{(2x + 1)^2 + 15}#

Now, performing the integration

#int frac{x^2}{x^2+x+4} "d"x = int "d"x - 1/2 int frac{2x+1}{x^2+x+4} "d"x #

#+ 7/2 int frac{1}{(x + 1/2)^2 + 15/4} "d"x#

#= x - 1/2 ln(x^2+x+4) + 7/sqrt15 tan^{-1}(frac{2x + 1}{sqrt15}) + c#

where #c# is the constant of integration.