Integral non-injective substitution ($u=x^2+1$) Integrate:
$$\int x^3 \sqrt{x^2+1} dx$$
My solution:

Choose $u=x^2+1 \Leftrightarrow x = \pm \sqrt{u-1}$, then du = $2x dx$. Therefore: $$\frac{1}{2} \int (u-1)(u)^{1/2} du = \frac{1}{2} \int (u^{3/2} - u^{1/2})du$$
$$= \frac{1}{5} (x^2+1)^{5/2} - \frac{1}{3}(x^2+1)^{3/2} + C$$

This is what I did and I am wondering. Because $x =\pm \sqrt{u-1}$, how do I know which, plus or minus, should I use? I just assumed to use the positive one because the function is in a power of $2$.
 A: Let's recall what a $u$ substitution is really doing in an indefinite integral. You are replacing $$\int f(x)dx $$ with $$ \int f(x(u))x'(u)du$$ and noting that if $\frac{d}{du}F(x(u)) = f(x(u))x'(u)$ then $F'(x) =f(x).$ In other words finding an anti-derivative for $f(x(u))x'(u)$ is the same as finding one for $f(x).$ 
Let's take your example slowly. Say you have $x(u) = \sqrt{u-1}.$ Then you look for an antiderivative of $f(\sqrt{u-1})\frac{1}{2\sqrt{u-1}}.$ Say you pick $x(u) =-\sqrt{u-1}.$ Then you look for an antiderivative of $-f(-\sqrt{u-1})\frac{1}{2\sqrt{u-1}}.$ Yikes! These are different functions so their antiderivatives will probably be different!
But it all has to work out in the end. Why? This is just using two different u-substitutions. There's no need for $x(u)$ to be the same thing. In fact many integrals have two different $u$ substitutions that work for it.
So your example actually gets your question backwards. The real issue is what if you want to substitute $x(u)= u^2+1$, in other words you want to use a non-invertable $x(u).$
Then trouble will come at the stage where you've found an antiderivative $G(u)$ for $ f'(x(u))x'(u)$ and you want to express it as $F(x(u))$. You need to know whether to plug in $u = +\sqrt{x(u)+1}$ or $u = -\sqrt{x(u)+1}$ into $G(u)$ in order to get $G(u)$ into the form $F(x(u)).$
Let's take the simplest example I can think of: $$ \int \frac{1}{2\sqrt{x}}e^{\sqrt{x}}dx.$$ We'd be inclined to use $x(u) = u^2$ here, so let's look at $$f(x(u))x'(u) = \frac{1}{2u}e^{u}2u = e^u.$$ Pretty easy to find an antiderivative... just $e^u$ (that's why we chose this $u$ substitution).
Now here's the tricky part. Naively, we would normally sub back in $u = \sqrt{x}$ and write the answer as $e^{\sqrt{x}} + C.$ But $x = u^2$ also admits another solution $u=-\sqrt{x}$ so we might be feel equally entitled to write $e^{-\sqrt{x}}+C,$ which would be wrong.
The key is you need to write the antiderivative $e^u$ as $F(x(u)) = F(u^2).$ We see that $e^u=e^{\sqrt{u^2}}$ is a valid way to do this and $e^{-\sqrt{u^2}}$ is not. So everything checks out, provided we're careful.
Generally, this issue never crosses one's mind  because you write $u = \sqrt{x}$ at the outset, cause it's 'what you're subbing for' in the function. So you know which branch of the inverse of $x(u)=u^2$ you chose.
Let's see what happens if we are inclined to choose differently. Consider if we instead had $$ \int -\frac{1}{2\sqrt{x}}e^{-\sqrt{x}}dx.$$ What if we wanted to "sub in $u=-\sqrt{x}$?" What we're really doing then is, again, taking $x(u) = u^2$ and getting $f(x(u))x'(u) = - e^{-u},$ antidifferentiating to obtain $e^{-u}$ and then realizing that's $e^{-\sqrt{u^2}}$ so that the anti derivative is $e^{-\sqrt{x}}$ How does this gel with "subbing in $u=-\sqrt{x}$?"  The minus sign that we got from integrating $e^{-u}$ instead of $e^u$ canceled with the minus sign from "du = -\frac{1}{2\sqrt{x}}dx". 
So there's actually no difference between subbing $u=\sqrt{x}$ and $u=-\sqrt{x}$ after all.
You can think of the 'u substitution ritual' as a convenient mnemonic for what's actually going on under the hood. (It doesn't help that it's completely backwards from what's going on, but it is much easier to use that way.)
