To solve the integral

$\int_{-1}^{7} \frac{dx}{\sqrt{2+x}},$

we can use the following steps:

Step 1: Substitution

Let's use the substitution $u = 2 + x$. Then, $du = dx$.

Step 2: Change of Limits

When $x = -1$, $u = 2 + (-1) = 1$. When $x = 7$, $u = 2 + 7 = 9$.

So the integral in terms of $u$ becomes:

$\int_{1}^{9} \frac{du}{\sqrt{u}}.$

Step 3: Simplification

The integral $\int \frac{du}{\sqrt{u}}$ can be simplified. We know that:

$\frac{1}{\sqrt{u}} = u^{-\frac{1}{2}}.$

Therefore, the integral becomes:

$\int u^{-\frac{1}{2}} \, du.$

Step 4: Integration

Integrate $u^{-\frac{1}{2}}$ with respect to $u$:

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

Step 5: Apply Limits

Now, we apply the limits 1 and 9 to the function $2\sqrt{u}$:

$\left[ 2\sqrt{u} \right]_{1}^{9} = 2\sqrt{9} - 2\sqrt{1} = 2(3) - 2(1) = 6 - 2 = 4.$

Final Answer

So, the value of the integral is:

$\boxed{4}.$

Would you like more details or have any questions?

Here are some related questions for further practice:

1. How would you evaluate $\int_{0}^{4} \frac{dx}{\sqrt{3+x}}$?
2. What if the integral was $\int_{0}^{1} \frac{dx}{\sqrt{1-x}}$?
3. How can you evaluate $\int_{1}^{4} \frac{dx}{(2+x)^2}$?
4. What is the integral of $\int \frac{dx}{\sqrt{x+5}}$ without definite limits?
5. How would you compute $\int_{1}^{8} \frac{dx}{(x-2)^{3/2}}$?

Tip: When dealing with integrals involving square roots, substitution often simplifies the problem significantly.