Integration rules: something so simple

[tsamocki]

I know the antiderivative of 1/(x^(1/2)) dx is 2(x^(1/2)) + constant, and i can prove it if i rewrite the inverse equation to be x^(-1/2). However, leaving it in the fraction form of "one divided by x to the one-half power", i always come up with 2/(3x^(3/2)).

Am i doing something wrong, or are there other rules for integration that i am unaware of?

Homework Statement

The two forms: the integral of x^(-1/2) dx and the integral of 1/(x^(1/2) dx.

Homework Equations

The formula for solving the most basic integrals is x^(n+1)/(n+1) plus a constant.

The Attempt at a Solution

In the fraction (inverse) form: the integral of 1/(x^(1/2) dx, i would solve it like this 1/(x^(1/2+1)/(1/2)+1; which gives me 1/(x^(3/2)/(3/2) = 2/((3x^(1/2)) + C.

However, in the form of x^(-1/2), i solve it by x^ ((-1/2) +1)/(-1/2)+1 = 2(x^(1/2).

Why is this the case?

 

[Dick]

There is no 'quotient rule'. The integral of 1/f(x) is not equal to 1/(the integral of f(x)). Treating it as x^(-1/2) is the correct way.