How do you find the most general antiderivative or indefinite integral #int (sqrtx/2+2/sqrtx)dx#?

1 Answer
Feb 20, 2018

Please see below.

Explanation:

If you do not distinguish the two, then use the power rule for integrals to get

#int (1/2x^(1/2) + 2x^(-1/2)) dx = 1/2 (2/3x^(3/2)) + 2(2/1x^(1/2))+C#

# = x^(3/2)/3+4x^(1/2)+C#

If (like Stewart) you consider the most general antiderivative to be the antiderivative on a maximal domain, the you must allow for a change of constants at the discontinuity of the integrand. (At #x=0#.) And the answer is

#f(x) = {(x^(3/2)/3+4x^(1/2)+C_1,"if",x < 0),(x^(3/2)/3+4x^(1/2)+C_2,"if",x > 0):}#