The case $\int x^2\,dx$ serves to illustrate a more general point, so I'll go out of my way to use $u$-substitution.

Select $u=x^2$. Then $$ \frac{du}{dx} = 2x \iff \frac{du}{2x}=dx\,. $$ Substituting we get $$ \int \frac{u}{2x}\,du\,. $$ But there's still an $x$ lingering. We can get rid of this by also substituting $x=\pm \sqrt{u}$: $$ \begin{split} \pm \int \frac{u}{2\sqrt{u}}\,du & = \pm \frac{1}{2} \int \sqrt{u}\,du\\ & = \pm \frac{u^{3/2}}{3}+c\\ & = \pm \frac{x^3}{3}+c \end{split} $$ So this still came out right. My question is: what's up with the $\pm$? I can't find a persuasive reason to just discard it.


Your last line is incorrect: you substituted "$\color{red}{x=\sqrt{u}}$" instead of $x=\pm\sqrt{u}$. Here's a corrected back substitution for $\pm u^{3/2}$: $$\pm u^{3/2}=\pm u^1\cdot u^{1/2}=\pm u\cdot\sqrt{u}=\pm x^2\cdot(\pm x)=+x^3.$$

UPDATE: By virtue of defining $u=x^2$, $u$ is automatically a non-negative quantity. Now, saying "$x=\pm\sqrt{u}$" does NOT define a function — instead, it defines two different functions, in the sense of being two different cases that we now have to consider.

Case 1: $x\ge0$. Then $x=+\sqrt{u}$ or equivalently $\sqrt{u}=+x$, and your answer (with the same "$+$" in front) is: $$+u^{3/2}=u^1\cdot u^{1/2}=u\cdot\sqrt{u}=x^2\cdot(+x)=x^3.$$

Case 2: $x<0$. Then $x=-\sqrt{u}$ or equivalently $\sqrt{u}=-x$, and your answer (with the same "$-$" in front) is: $$-u^{3/2}=-u^1\cdot u^{1/2}=-u\cdot\sqrt{u}=-x^2\cdot(-x)=x^3.$$

So it's $x^3$ either way.

  • $\begingroup$ What I did in my mind at least was $u^{3/2} = (x^2)^{3/2}=x^3$. That would be a completely valid derivation under any other circumstances. How can a $\pm$ out front affect that? $\endgroup$ – Sebastian Oberhoff May 12 '18 at 3:54
  • $\begingroup$ @SebastianOberhoff: No, it wouldn't. For example, evaluate $x^3$ and $(x^2)^{3/2}$ when $x=-1$. $\endgroup$ – zipirovich May 12 '18 at 5:17
  • $\begingroup$ That's a good point. But I'm still suspicious of your reasoning. It just seems so reverse engineered to me. I don't see any way to extract the essence of what you're saying in a way that would allow me to arrive at the right answer without knowing what it ought to be beforehand. $\endgroup$ – Sebastian Oberhoff May 12 '18 at 5:24
  • $\begingroup$ @SebastianOberhoff: See the update in my answer above. $\endgroup$ – zipirovich May 12 '18 at 5:33
  • $\begingroup$ "$u=\pm\sqrt{x}$" needs to be changed to "$x=\pm\sqrt{u}$" but I can't submit edits less than 6 characters in length. $\endgroup$ – Sebastian Oberhoff May 12 '18 at 5:40

Instead of $u=x^2$ which is problematic, you better let $$x= \sqrt u, dx= \frac {du}{ 2\sqrt u}$$

Then there is no problem with the integral.$$ \int x^2 dx = (1/2)\int \sqrt u du = (1/3)u^{3/2}+C =(1/3)x^3+C $$


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