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.

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 : $$=\left[\frac{\left(x+2\right)^{2.5}}{2.5}-\frac{2\left(x+2\right)^{1.5}}{1.5}\right]^1_0$$
$$=\left[\frac{\left(1+2\right)^{2.5}}{2.5}-\frac{2\left(1+2\right)^{1.5}}{1.5}\right]-\left[\frac{\left(0+2\right)^{2.5}}{2.5}-\frac{2\left(0+2\right)^{1.5}}{1.5}\right]$$
$$=-4\sqrt{3}+\frac{16\sqrt{2}+54\sqrt{3}}{15}$$
