The given mathematical expression in the image is:

$\int \frac{dx}{3 - \sqrt{x}}$

Solution:

To solve this integral, we will use substitution techniques.

Step 1: Substitution

Let: $u = 3 - \sqrt{x}$ Then: $\sqrt{x} = 3 - u \quad \text{and} \quad x = (3 - u)^2$ Differentiating $x = (3 - u)^2$: $dx = 2(3 - u)(-1) \, du = -2(3 - u) \, du$

Step 2: Rewrite the Integral

Substitute $u = 3 - \sqrt{x}$ and $dx = -2(3 - u) \, du$ into the integral: $\int \frac{dx}{3 - \sqrt{x}} = \int \frac{-2(3 - u) \, du}{u}$

Simplify: $= -2 \int \frac{3 - u}{u} \, du = -2 \int \left( \frac{3}{u} - 1 \right) \, du$

Step 3: Solve the Integral

Split the integral: $-2 \int \left( \frac{3}{u} - 1 \right) \, du = -2 \left( 3 \int \frac{1}{u} \, du - \int 1 \, du \right)$

Solve each term: $\int \frac{1}{u} \, du = \ln|u|, \quad \int 1 \, du = u$

Thus: $-2 \left( 3 \ln|u| - u \right) = -6 \ln|u| + 2u$

Step 4: Back-Substitute $u$

Recall $u = 3 - \sqrt{x}$: $-6 \ln|u| + 2u = -6 \ln|3 - \sqrt{x}| + 2(3 - \sqrt{x})$

Simplify: $-6 \ln|3 - \sqrt{x}| + 6 - 2\sqrt{x}$

Final Answer:

$\int \frac{dx}{3 - \sqrt{x}} = -6 \ln|3 - \sqrt{x}| + 6 - 2\sqrt{x} + C$

Would you like more clarification or details on any step?

Related Questions:

1. What are the key substitution techniques used in integrals involving square roots?
2. How do you handle logarithmic terms in definite integrals?
3. What other functions require substitutions like $u = a - \sqrt{x}$?
4. How can this solution be checked by differentiation?
5. What is the significance of absolute values in logarithmic terms?

Tip:

For integrals involving terms like $a - \sqrt{x}$, substitution with $u = a - \sqrt{x}$ often simplifies the problem significantly.