To solve the integral 

\[
\int \frac{x + 3}{\sqrt{x^2 + 6x}} \, dx,
\]

we can use a substitution to simplify the integrand.

### Step 1: Simplify the Expression
First, let's rewrite the integrand for clarity:

\[
\frac{x + 3}{\sqrt{x^2 + 6x}}.
\]

Notice that the expression under the square root, \(x^2 + 6x\), can be completed to form a perfect square:

\[
x^2 + 6x = (x^2 + 6x + 9) - 9 = (x + 3)^2 - 9.
\]

So, the integral becomes:

\[
\int \frac{x + 3}{\sqrt{(x + 3)^2 - 9}} \, dx.
\]

### Step 2: Use Substitution
Let:

\[
u = x + 3.
\]

Then,

\[
du = dx, \quad x = u - 3.
\]

The integral now becomes:

\[
\int \frac{u}{\sqrt{u^2 - 9}} \, du.
\]

### Step 3: Further Simplify Using Another Substitution
Now, consider the substitution:

\[
v = \sqrt{u^2 - 9}.
\]

Then,

\[
v^2 = u^2 - 9, \quad u^2 = v^2 + 9, \quad u = \sqrt{v^2 + 9}.
\]

Differentiating with respect to \(v\):

\[
du = \frac{v}{\sqrt{v^2 + 9}} \, dv.
\]

The integral becomes:

\[
\int \frac{\sqrt{v^2 + 9}}{v} \cdot \frac{v}{\sqrt{v^2 + 9}} \, dv = \int dv = v.
\]

### Step 4: Back-substitute to Original Variable
Recall that:

\[
v = \sqrt{u^2 - 9} = \sqrt{(x + 3)^2 - 9} = \sqrt{x^2 + 6x}.
\]

Thus, the integral evaluates to:

\[
\sqrt{x^2 + 6x} + C,
\]

where \(C\) is the constant of integration.

### Final Answer
\[
\int \frac{x + 3}{\sqrt{x^2 + 6x}} \, dx = \sqrt{x^2 + 6x} + C.
\]

Would you like more details on any of these steps, or do you have any questions?

### Related Questions:
1. How do you complete the square for expressions in integrals?
2. What other types of substitutions can simplify integrals involving square roots?
3. Can trigonometric substitutions be used for this integral, and how?
4. What is the geometric interpretation of the integral of \(1/\sqrt{x^2 + ax}\)?
5. How does changing the variable affect the limits in a definite integral?

### Tip:
When you see an expression like \(x^2 + ax\) under a square root, consider completing the square to simplify the integrand.

Learn how to solve the integral ∫ (x+3) / (x² + 6x)^(1/2) dx using substitution and completing the square. This detailed guide walks you through each step, making it suitable for students in advanced high school or college calculus.

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