We are tasked with evaluating the definite integral

$\int_{1}^{9} \frac{1}{\sqrt{x} (2 + \sqrt{x})^2} \, dx$

by substitution.

Step 1: Substitution

Let’s perform a substitution to simplify the integrand. We can set:

$u = \sqrt{x}$

Thus, $x = u^2$ and $dx = 2u \, du$. The limits of integration also change accordingly:

• When $x = 1$, $u = \sqrt{1} = 1$,
• When $x = 9$, $u = \sqrt{9} = 3$.

Now the integral becomes:

$\int_{1}^{3} \frac{1}{u (2 + u)^2} \cdot 2u \, du$

Step 2: Simplify

Notice that $\frac{1}{u} \cdot 2u = 2$. The integral reduces to:

$\int_{1}^{3} \frac{2}{(2 + u)^2} \, du$

Step 3: Solve the Integral

Now, we solve this integral. We can apply the substitution:

$v = 2 + u \quad \text{so that} \quad dv = du$

The limits of integration change as well:

• When $u = 1$, $v = 2 + 1 = 3$,
• When $u = 3$, $v = 2 + 3 = 5$.

Thus, the integral becomes:

$\int_{3}^{5} \frac{2}{v^2} \, dv$

This is a standard integral. We know that:

$\int \frac{1}{v^2} \, dv = -\frac{1}{v}$

Therefore, the integral becomes:

$-2 \left[ \frac{1}{v} \right]_{3}^{5}$

Step 4: Compute the Definite Integral

Now, evaluate the expression at the limits:

$-2 \left( \frac{1}{5} - \frac{1}{3} \right) = -2 \left( \frac{3 - 5}{15} \right) = -2 \left( \frac{-2}{15} \right) = \frac{4}{15}$

Thus, the value of the definite integral is:

$\boxed{\frac{4}{15}}$

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Would you like more details on any part of this solution, or do you have any questions?

Here are 5 related questions to deepen understanding:

1. How would the result change if we used different limits of integration?
2. Can you derive the result of the indefinite integral before applying the limits?
3. How does substitution help simplify integrals in general?
4. What would happen if we had a different power on the term $(2 + \sqrt{x})$?
5. Can you explain how to handle integrals that involve more complex substitutions?

Tip: When choosing a substitution, look for parts of the integrand that resemble the derivative of other parts—it simplifies the process!