I have the following integrals $$\int x\sqrt{x-1}\, dx, \quad \text{and} \quad \int \sqrt{e^x-1}\,dx$$ I am not interested in solving integrals but rather in a consideration that I have thought when I have solved the integrals to my high school students. I generally put the positions in such a way that there is no double sign. In other words I would place in the first case $x-1=t$ and in the second case $e^x-1=t$. I have found that in the adopted textbook, in the first case, the position is adopted instead is: $$x-1=t^2$$ In the second case however I can use for analogy ($e^x-1=t^2$) It is true that $x-1>0$, and $t^2>0$, but if $$x-1=t^2 \implies t=\pm \sqrt{x-1}$$
Why would I choose the $+$ sign instead of the minus sign when I operate substitutions under square root? Is there a specific reason?
The answer will be the same regardless of whether you choose the $+$ or $-$ sign.
Let's say you want to integrate $\int \sqrt{f(x)} \, dx$. We can write $t^2 = f(x) \implies t = \pm \sqrt{f(x)} \implies \pm t = \sqrt{f(x)}$.
Case $1$: You choose $t = \sqrt{f(x)}$. $$\frac{dt}{dx} = g(t) \implies dx = \frac{dt}{g(t)}$$ $$\therefore \int \sqrt{f(x)} \, dx = \int t \, \frac{dt}{g(t)}$$
Case $2$: You choose $-t = \sqrt{f(x)} \implies t = -\sqrt{f(x)} $ $$\frac{dt}{dx} = -g(t) \implies dx = -\frac{dt}{g(t)}$$ $$\therefore \int \sqrt{f(x)} \, dx = \int -t \cdot \left(\, -\frac{dt}{g(t)} \right)= \int t \, \frac{dt}{g(t)}$$
As demonstrated, in both cases, you get the same result.