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In general, why don't we consider the negative case when doing $u$ substitutions that involve a $u^2$ (i.e. when subbing back in for $x$ the square root only has the positive value $+\sqrt{x}$ rather than $-\sqrt{x}$ or whatever function is being square rooted)

A simple example from my textbook:

integrate $f(x)=\int \frac{1}{1+\sqrt{x}} dx $ by substitution of $x=(u-1)^2$

through some simple substitution the textbook answers reach:

$f(x)=2(1+\sqrt{x})-2(\ln(1+\sqrt{x}))+C$

why cannot the case $f(x)=2(1-\sqrt{x})-2(\ln(1-\sqrt{x}))+C$ work in this method?

Is it just that you first define $u=1+\sqrt{x}$ rather than the substitution they give in the question, and then reach the $x(u)$ function?

I thought that it might be because $x(u)$ and $u(x)$ functions must be inverse of each other, but you can still define that inverse as the negative and it should work. Any input is appreciated, am I just missing something extremely obvious?

tgsweat
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    The substitution is equal $u=1+\sqrt{x}.$ Then you solve for $x$ to get $x=(u-1)^2.$ – Ryszard Szwarc Jun 04 '22 at 06:07
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    Have you tried differentiating the answer with the minus sign to see if you get back the integrand? – Gerry Myerson Jun 04 '22 at 06:09
  • @GerryMyerson I understand that it's not correct from the integration standpoint, but using the method as given it appears as if it should. – tgsweat Jun 04 '22 at 06:16
  • @RyszardSzwarc I thought it was this, it seems to make the most sense. When doing a similar thing with trigonometric functions we also assume it's in the quadrant of the inverse function, which was why i guessed $x(u)$ and $u(x)$ just must be inverse, but it appears the simplest answer is the right one. Thanks! – tgsweat Jun 04 '22 at 06:19
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    Please refer to my answers here and here. The short of it is that an implicit substitution (that is, where the substitution involves a function $h(u)$ of the new variable $u)$ does require $h$ to be invertible, so it may need to be restricted: in your case, choosing either $u\ge1$ or $u\le1$ will work. – ryang Jun 04 '22 at 09:27

2 Answers2

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Let $u = 1 - \sqrt{x}$. Then $x = (u - 1)^2$, so $dx = 2(u - 1) du$.

And $\sqrt{x} = \sqrt{(u - 1)^2} = |u - 1|$. Note that $u \le 1$, so $u - 1 \le 0$, so $|u - 1| = -(u - 1) = 1 - u$.

$$\int \frac{1}{1 + \sqrt{x}} dx$$ $$\int \frac{1}{1 + ( 1 - u)} 2(u-1) du$$ $$\int \frac{2u - 2}{2 - u} du$$ $$\int \frac{2 - 2(2 - u)}{2 - u} du$$ $$\int (\frac{2}{2 - u} - 2) du$$ $$-2 \ln{|2 - u|} - 2u + C$$ $$-2 \ln{|2 - (1 - \sqrt{x})|} - 2(1 - \sqrt{x}) + C_1$$ $$-2 \ln{|1 + \sqrt{x}|} - 2 + 2\sqrt{x} + C_1$$ $$2\sqrt{x} - 2 \ln(1 + \sqrt{x}) + C_2$$

Which is the same as your textbook gets for the $u = 1 + \sqrt{x}$ substitution, except differing by a constant (which is irrelevant anyway for indefinite integrals).

How did you get your answer? You forgot the absolute value operators, didn't you?

Dan
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  • That wasn't 'my answer'

    I also got what you have provided here, I know how to do this substitution, it's trivial.

    The textbook asks you to use $x=(u-1)^2$ rather than $u=1+\sqrt{x}$ which is fundamentally the same thing, but for the textbook-provided substitution for $x(u)$ rather than $u(x)$ (as both you and I did) $u$ is equated to $u=1 +/- \sqrt{x}$

    When substituting back for $u$ at the end, this yields a possibility for the $1-\sqrt{x}$ option, which is incorrect and hence why I asked the question.

    My question was not how to integrate it but why the +ve root is used.

     – tgsweat Jun 04 '22 at 06:41
  • Apologies, new to this so I'm not sure of the +/- notation. – tgsweat Jun 04 '22 at 06:42
  • I have read through your response more clearly, and it makes more sense the point you're making, thanks! Ignore my last two comments, mentally resolved now :) – tgsweat Jun 04 '22 at 06:48
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When making a substitution of the form $x=g(u)$, it must be the case that $g$ is a differentiable one-to-one function. Thus, when we substitute $x=(u-1)^2$, it is implicit that either $u$ is always greater than $1$, or $u$ is always less than $1$. If $u>1$, then $\sqrt{x}=u-1$, and we arrive at your textbook's solution. If $u<1$, then $\sqrt{x}=1-u$, and $dx=2(u-1) \, du$, so \begin{align} I&=\int\frac{2(u-1)}{2-u} \, du \\[5pt] &=-2(u+\log|u-2|)+K \\[5pt] &=-2(1-\sqrt{x}+\log|-\sqrt{x}-1|)+K \\[5pt] &=-2\left(1-\sqrt{x}+\log\left(1+\sqrt{x}\right)\right)+K \, , \\[5pt] \end{align} which is the same as the textbook's solution (set $K=C+4$).

Joe
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