Choosing a substitution to evaluate $\int \frac{x+3}{\sqrt{x+2}}dx$

Question

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?

Answer 1

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

$$I = \frac{2}{3}\sqrt{x+2}(x+5)$$

Answer 2

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 $$\int \frac{du}{\sqrt{u}} = 2 u^{1 / 2} + C'.$$

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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): $$\int \frac{(v^2 - 2) + 3}{v} (2v) \, dv = 2 \int (v^2 + 1) \,dv .$$

Answer 3

Let $\sqrt{x+2}=t\implies \frac{dx}{2\sqrt{x+2}}=dt$ or $dx=2t\ dt$ $$\int \frac{x+3}{\sqrt{x+2}}\ dx$$$$=\int \frac{t^2-2+3}{t}(2t\ dt)$$ $$=2\int (t^2+1)\ dt$$ $$=2\left(\frac{t^3}{3}+t\right)+C$$ $$=2\left(\frac{(x+2)^{3/2}}{3}+\sqrt{x+2}\right)+C$$ $$=\frac 23(x+5)\sqrt{x+2}+C$$