#int(x^3/(x + 2))# equals ??

1 Answer
Apr 29, 2018

#x^3/3-x^2+4x-8ln|x+2|+C#

Explanation:

We can divide out the integrand using

#x^3/(x+2) = (x^3+2^3-2^3)/(x+2) = x^2-2x+4-2^3/(x+2)#

Thus

#int x^3/(x+2)dx = int\ [x^2-2x+4-8/(x+2)]dx#
#qquad qquad qquad qquad quad = x^3/3-x^2+4x-8ln|x+2|+C#