Don't understand integration using a substitution rule for a square root function. As an example, evaluate $(1) \int \sqrt{2x + 1} \mathrm{d}x$.
One way would be to let $u = 2x+1$, and I do understand that one as it clearly fits with my conception of the substitution rule as a sort of "reverse chain rule" for antiderivatives - this choice of $u$ leads to $\mathrm{d}u=2\mathrm{d}x$, which then needs to be corrected by a multiplication of $\frac{1}{2}$ for the missing 2 in (1). 
However, another possible substitution is apparently $u= \sqrt{2x+1}$, which leads to $\mathrm{d}u=\frac{\mathrm{d}x}{\sqrt{2x+1}}$.
I don't get this at all. If the substitution rule is $$\int f(g(x))g'(x)\mathrm{d}x = \int f(u)\mathrm{d}u$$ with $u=g(x)$, then what exactly is $g'(x)$ and what is $f(x)$ supposed to be in the latter substitution approach? And what is the correction factor? I just can't wrap my head around it (I'm a beginner, so please keep that in mind!)
 A: With $f(u)=u^2$ and $g(x)=\sqrt{2x+1}$, the substitution rule gives
$$
\int u^2 \, du
= \int f(u) \, du
= \int f(g(x)) \, g'(x) \, dx
= \int (\sqrt{2x+1})^2 \, \frac{2}{2\sqrt{2x+1}} \, dx
= \int \sqrt{2x+1} \, dx
.
$$
Now read this chain of equalities from right to left, to get it in the order that you would normally write the calculation:
$$
\int \sqrt{2x+1} \, dx
= \dots
= \int u^2 \, du
= \frac{u^3}{3} + C
= \frac{(\sqrt{2x+1})^3}{3} + C
= \frac{(2x+1)^{3/2}}{3} + C
.
$$
A: We have:
\begin{align*}
f(x) &= x^2 \\
g(x) &= \sqrt{2x + 1} \\
g'(x) &= \frac{1}{\sqrt{2x + 1}}
\end{align*}
which leads to:
\begin{align*}
\int \sqrt{2x + 1} ~\mathrm{d}x
&= \int (\sqrt{2x + 1})^2 \cdot \frac{1}{\sqrt{2x + 1}} ~\mathrm{d}x \\
&= \int u^2 ~\mathrm{d}u
\end{align*}

Remark: Observe that if $u = \sqrt{2x + 1}$, then:
\begin{align*}
u^2 &= 2x + 1 \\
2u~\mathrm{d}u &= 2~\mathrm{d}x \\
u~\mathrm{d}u &= \mathrm{d}x
\end{align*}
So by replacing $\sqrt{2x + 1}$ with $u$ and by replacing $\mathrm{d}x$ with $u~\mathrm{d}u$, we obtain:
$$
\int \sqrt{2x + 1} ~\mathrm{d}x
= \int (u) (u~\mathrm{d}u)
= \int u^2 ~\mathrm{d}u
$$
A: You are correct the initial idea behind the substitution integrals is the reverse of the chain rule. We look for a composition of two functions which is clear here it is $f=\sqrt{x}$ and $g= 2x+1$. Or we could have $g = \sqrt{2x+1}$ and $f=x^2 $
$$\int \sqrt{2x+1 } dx = \int \underbrace{\frac{2}{\sqrt{2x+1}}}_{g'} \underbrace{\left(\sqrt{2x+1}\right)^2}_{f(g)} dx $$ 
But when you master the idea of the integral of substitution sometimes its going to be hard to guess what is $g$ and $f$ are for example 
$$\int \frac{x^5}{\sqrt{x^3+1}} dx $$ here we usually go by letting $u$ be the inside of the square i.e. the function which is inside another function. Here if you try it you will see that $du = 3x^2 dx$ then $x^2$ will cancel to have $x^3$ in top which can be written in terms of $u$. So in future what you will do is try to guess a good $u$ such that when you eliminate $x$ you will get something we can integrate easily. What am trying to say you should think about the substitution method as writing the function in way where we can integrate. I hope that helped you 
A: You're on the right track!
Let $u= \sqrt{2x+1}$. This leads to $\mathrm{d}u=\frac{\mathrm{d}x}{\sqrt{2x+1}}$, thus $ \sqrt{2x+1}\mathrm{d}u=\mathrm{d}x$. By definition of $u$, we have $ u\mathrm{d}u=\mathrm{d}x$.
Now we substitute $u$ and $\mathrm{d}u$ into the integral as follows
$$ \int \sqrt{2x+1}\mathrm{d}x= \int u \mathrm{d}x = \int u (u\mathrm{d}u) = \int u^2 \mathrm{d}u = \frac{1}{3}u^3+C = \frac{1}{3}(\sqrt{2x+1})^3+C$$
