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?
 A: let's make it easier than that!
Use this:
$$x + 2 = t^2 ~~~~~~~~~~~ x+3 = t^2 + 1 ~~~~~~~ \text{d}x = 2t\ \text{d}t$$
Obtaining
$$I = \int\frac{t^2 + 1}{t}\ 2t\ \text{d}t = 2\int t^2 + 1\ \text{d}t =  \frac{2}{3}t^3 + 2t$$
Coming back to $x$, having $t = \sqrt{x+2}$ and you'll have
$$I = \frac{2}{3}\sqrt{x+2}(x+5)$$
A: We don't need to apply the second substitution (in fact, it is circular): Using the general rule $\int u^m = \frac{1}{m + 1} u^{m + 1}$ (for $m \neq -1$), we have
$$\int \sqrt{u} \,du = \int u^{1 / 2} du = \frac{2}{3} u^{3 / 2} + C$$ and likewise $$\int \frac{du}{\sqrt{u}} = 2 u^{1 / 2} + C'.$$

On the other hand, we could instead at the first step make the rationalizing substitution $$v = \sqrt{x + 2},$$ so that $x = v^2 - 2$ and hence $dx = 2 v \,dv$. This has the advantage that the resulting integral expression is rational (in fact, in this case, polynomial):
$$\int \frac{(v^2 - 2) + 3}{v} (2v) \, dv = 2 \int (v^2 + 1) \,dv .$$
A: 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.
A: Let $\sqrt{x+2}=t\implies \frac{dx}{2\sqrt{x+2}}=dt$ or $dx=2t\ dt$
$$\int \frac{x+3}{\sqrt{x+2}}\ dx$$$$=\int \frac{t^2-2+3}{t}(2t\ dt)$$
$$=2\int (t^2+1)\ dt$$
$$=2\left(\frac{t^3}{3}+t\right)+C$$
$$=2\left(\frac{(x+2)^{3/2}}{3}+\sqrt{x+2}\right)+C$$
$$=\frac 23(x+5)\sqrt{x+2}+C$$
