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.
 A: It's possible to let $x = t^2-2$ and we all agree that $t = \pm\sqrt{2}$ and $t = \pm\sqrt{3}$ are solutions, but saying
$$\int_{\pm\sqrt{2}}^{\pm\sqrt{3}}t^4-2t^2dt$$
is meaningless. An integral bound shouldn't have two different numbers in it.
The substitution should be monotonic, which basically means it should vary in such a way that it either never decreases, or vary in such a way that it never increases.  In this case, if you really want to let $t^2 = x+2$, it means either $t = \sqrt{x+2}$ or $t = -\sqrt{x+2}$ is one of your options. Notice how those functions are monotonic because the first one is always increasing and the second one is always decreasing.
If you choose the first option, then the lower and upper bound will be $\sqrt{2}$ and $\sqrt{3}$, respectively, which results in
$$2\int_{\sqrt{2}}^{\sqrt{3}}\left(t^{4}-2t^{2}\right)dt.$$
Alternatively, if you choose the second option, then the lower and upper bound will be $-\sqrt{2}$ and $-\sqrt{3}$, respectively, which results in
$$-2\int_{-\sqrt{2}}^{-\sqrt{3}}\left(t^{4}-2t^{2}\right)dt.$$
A: Given that : $$=\int _0^1\:\left(x\sqrt{x+2}\right)dx$$
Let : $$u=x+2$$ $$\frac{du}{dx}=1$$ $$du=dx$$
Also : $$u=x+2$$ $$x=u-2$$
Substitute into the equation : $$=\int \left(\left(u-2\right)\sqrt{u}\right)du$$
Then : $$=\int \left(u^{1.5}\:-2u^{0.5}\right)du$$
Integrate : $$=\frac{u^{2.5}}{2.5}-\frac{2u^{1.5}}{1.5}$$
Substitute into $x$ back and put the original limit : $$=\left[\frac{\left(x+2\right)^{2.5}}{2.5}-\frac{2\left(x+2\right)^{1.5}}{1.5}\right]^1_0$$
$$=\left[\frac{\left(1+2\right)^{2.5}}{2.5}-\frac{2\left(1+2\right)^{1.5}}{1.5}\right]-\left[\frac{\left(0+2\right)^{2.5}}{2.5}-\frac{2\left(0+2\right)^{1.5}}{1.5}\right]$$
$$=-4\sqrt{3}+\frac{16\sqrt{2}+54\sqrt{3}}{15}$$
A: 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 $t\geq0$.
