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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 $2$.

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    hint :$$x=\sinh u \to \sqrt{x^2+1}=\sqrt{(\sinh u)^2+1}= \cosh u \\ \int (\sinh u )^3 \cosh u (\cosh u du)$$ – 2017-02-09
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    You didn't let $u = \pm \sqrt{x^2+1},$ you let $u = x^2+1$ and so solving for $x$ gave $\pm \sqrt{u-1}$ – 2017-02-09
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    Typo. My solution is correct to the above person. I don't know what you are doing. – 2017-02-09
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    Finding an antiderivative is, in most cases, a very informal procedure. In order to be more precise, the problem itself should specify a domain for the integrand and it does not. So pick whatever you want. – 2017-02-09
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    @GitGud Wouldn't the negative sign change the integral, or vice versa? Not this one because its in the power of 2 but like a different one – 2017-02-09
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    @user349557 It shouldn't, the minus sign would carry through with the derivative too, eventually canceling. In any case the theory behind the substitution is, I think, more enlightening than thinking about such examples. See, for instance, [this](https://en.wikipedia.org/wiki/Integration_by_substitution#Definition). What you did is simply present two choices for $\phi$. There's no mention of $\phi$ being unique and you've probably seen a few examples where several substitutions work. This is another one. – 2017-02-09
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    Thank you, I understand now – 2017-02-09
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    If $u=x^2+1$, then $\text{d}u=2x\text{d}x$. Now, $$x^3\sqrt{x^2+1}\text{d}x=\frac{1}{2}x^2\sqrt{x^2+1}\cdot 2x\text{d}x=\frac{1}{2}(u-1)\sqrt{u}\text{d}u$$ Notice that you don't even need to express $x$ in terms of $u$. – 2017-02-09

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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.)

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    Thx for writting all this – 2017-02-09