Doubts in solving $\int_0^1 x\sqrt{x+2} dx$ I was solving the following integral and got stuck in finding the new limits after the substitution. $$\int_0^1 x\sqrt{x+2} dx$$

Here's my work so far:
Putting $(x+2) = t^2$ so that $dx = 2t\ dt$ and $x = t^2 - 2$
Thus the original integral changes to,
$$\int_{?}^{?} (t^2 - 2) (t) (2t )\ dt = \boxed{2\int_{?}^{?} t^4 - 2t^2\ dt}$$
The above boxed integral can be solved easily using power rule of integral but I'm totally out in finding the limits.
When $x =0, t^2= 2$ and when $x = 1, t^2 = 3$ so $t$ is going from $\pm \sqrt{2}$ to $\pm \sqrt{3}$.
So we need to evaluate the integral $$2\int_{t = \pm \sqrt{2}}^{t = \pm \sqrt{3}} t^4 - 2t^2\ dt$$
which is not meaningful I think.
So, I need help in finding the sign convention of the limits of the integral.

Here’s my take on the problem: When you write $t^2=x+2$ and then proceed to substitute $\sqrt{x+2}$ with $t$, then you are essentially writing that $$\sqrt {t^2}=t$$, implying that $$|t|=t$$ which means that $t\geq0$. Thus I would prefer the limits be from $\sqrt2$ to $\sqrt3$.
As a side note, the problem can be circumvented by assuming $t=\sqrt{x+2}$, which is an equivalent substitution but easily tells us that $t\geq0$.
