# 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?

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

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

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

• How did you come up with this solution? To me, it looks just incredibly clever; is there a specific rule to identify this kind of cases? – Johnny Bueti Jan 31 '16 at 12:15
• Very often if an integrand involves a radical, substituting by setting a new variable to be the radical expression improves the situation, and often will turn algebraic integrands into rational ones, which can then be handled with the Method of Partial Fractions and standard integrals one already knows. For example, one can integrate $\int \sqrt{\frac{1-x}{1+x}} \,dx$ by first substituting $u = \sqrt{\frac{1-x}{1+x}}$. Rearranging gives that $x$ is a rational expression in $u$, and hence so is $dx$. (I recommend this as an exercise, by the way, as it illustrates the technique nicely.) – Travis Willse Jan 31 '16 at 12:38
• Perhaps the downvoter would explain their objection? – Travis Willse Jan 31 '16 at 12:39
• Thank you. I tried to substitute $u = \sqrt{\frac{1-x}{1+x}}$, differentiate and explicit the $x$ value in respect to $u$ to substitute it in the $u$ differential and get the new integral written entirely in respect to $u$, but it looks... well: $\int {-\frac{1}{1 + \left ( \frac{1-u^2}{1+u^2} \right )^2}du}$ – Johnny Bueti Jan 31 '16 at 12:58
• I think there may be an error in your simplification, but as it stands, we can multiply the numerator and denominator of the integrand by $(1 + u^2)^2$ and expand, giving $-\int\frac{(1 + u^2)^2}{(1 + u^2)^2 + (1 - u^2)^2} du$. – Travis Willse Jan 31 '16 at 15:14

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.

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

• Isn't $dt = \frac{1}{2\sqrt{x+2}}dx$ ? – Johnny Bueti Jan 31 '16 at 13:00
• yes, you are right, but substituting $\sqrt{x+2}=t$ one should get $dx=2t\ dt$ – Harish Chandra Rajpoot Jan 31 '16 at 13:22