How do you find the integral of #(x^4+x-4) / (x^2+2)#?

1 Answer
May 19, 2015

First, use polynomial long division to get #(x^4+x-4)/(x^2+2)=x^2-2+x/(x^2+2)#. Next, integrate this expression, using a substitution #u=x^2+2, du=2x\ dx# to get:

#\int(x^4+x-4)/(x^2+2)dx=\int(x^2-2+x/(x^2+2))dx#

#=\frac{1}{3}x^3-2x+\frac{1}{2}ln(x^2+2)+C#

Note that absolute value signs are not needed in the argument of the logarithm function since #x^2+2>0# for all real #x#.