I have trouble understanding when I can simply take the anti-derivative of square root functions and when I have to solve by other means. I don't know when it is and when it is not okay to simply take the anti-derivative of a square root function when I'm trying to find the integral. Could someone explain to me when and when it is not appropriate? (Examples below of what I mean since my question sounds a bit unclear)
Example 1: $\int x^{1/2}~dx = \frac{2}{3} x^{3/2} + C$
Example 2: $\int (2x+1)^{1/2}~dx = \frac{1}{3} (2x+1)^{3/2} + C$
In cases 1 and 2, it was perfectly okay for me to simply take the anti-derivative of the square root function. My question is: why or why not can I apply this same logic to the following examples (ex.3 and 4)? How should I solve the following problems then, and what kinds of generalizations can I make about solving integrals of square root functions? 
Example 3: $\int (2x^4+1)^{1/2}~dx = \frac{1}{12x^3} (2x^3+1)^{3/2} + C$
Example 4: $\int (1+e^{-2x})^{1/2}~dx = \frac{-1}{3e^{-2x}} (1+e^{-2x})^{3/2} + C$
Thank you so much for your help! Also was hoping to apply the case to indefinite and definite integrals
 A: The main difference between your two groups of examples is that in the first two, the stuff under the square root is just a linear polynomial; in the second two, it has more complicated functions. When you have higher polynomials under the square roots, regular substitution won't suffice (even though it did in your second example), unless the square root is multiplied by something convenient on the outside, as in the case of $\int 2x\sqrt{x^2+1}dx$. Often you'll have to use trigonometric substitution, or even something with a bit more ingenuity. 
A: If you think about these exercises in terms of substitutions, perhaps it will be clearer. For example 2, if you set $u = 2x+1$, then $du = 2\,dx$, or $dx = \frac{1}{2}\,du$, so that
$$\int (2x+1)^{1/2}\,dx = \frac{1}{2}\int u^{1/2}\,du = \frac{1}{3}u^{3/2}+C = \frac{1}{3}(2x+1)^{3/2}+C.$$
But try that with example 3: if $u = 2x^4+1$, then $du = 8x^3\,dx$, and there is no simple expression for the integral in terms of $u$.
So the only reason example 2 works is that $du$ is a constant multiple of $dx$. There are other situations where this substitution will work, but certainly it does not work in all cases.
