Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$ Is there any other value you can assign to the substitution variable to solve this integral?

$$\int \frac{x+3}{\sqrt{x+2}}dx$$

Substituting $u = x + 2$:
$$du = dx; u +1 = x+3 ,$$
and we get this new integral that we can then split into two different ones:
$$\int \frac{u + 1}{\sqrt{u}}du = \int \frac{u}{\sqrt{u}}du + \int \frac{1}{\sqrt{u}}du .$$
We can substitute again $s = \sqrt u$ and get two immediate integrals:
$$s = \sqrt{u}; \quad ds = \frac{1}{2\sqrt{u}}du; \quad 2s^2 =u .$$ Substituting back $u$ to $s$ and $x$ to $u$ we get this result,
$$s^2 + \ln{\left | \sqrt{u} \right |} = u + \ln{\left | \sqrt{u} \right |} = x+2+\ln{\left | \sqrt{x+2} \right |},$$ which doesn't look quite to be right. What am I doing wrong? I'm pretty unsure about the second substitution, $2s^2 = u$. Is it correct?

An other way is to write 
$$\int\frac{x+3}{\sqrt{x+2}}dx=\int\frac{x+2+1}{\sqrt{x+2}}dx$$
$$=\int\frac{x+2}{\sqrt{x+2}}dx+\int\frac{1}{\sqrt{x+2}}dx=\int\sqrt{x+2}dx+\int\frac{1}{\sqrt{x+2}}dx$$
$$=I_1+I_2.$$
In $I_1$ we put the change of variable $u=x+2,\ du=dx$ and in $I_2$ we put $w=\sqrt{x+2},\ dw=\frac{1}{2\sqrt{x+2}}dx$, after calculations we obtain
$$\int\frac{x+3}{\sqrt{x+2}}dx=\frac{2}{3}\left(x+2\right)^{1/2}(x+5)+C,$$ as mentioned in the above result.
