The integral presented in the image is:

$\int \frac{\sqrt{x} + 2}{\sqrt{x} - 1} \, dx$

Solution:

To solve this integral, we will simplify and decompose it:

1. Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u \, du$. Substituting into the integral:

   $\int \frac{\sqrt{x} + 2}{\sqrt{x} - 1} \, dx = \int \frac{u + 2}{u - 1} \cdot 2u \, du$

   This simplifies to:

   $2 \int \frac{u(u + 2)}{u - 1} \, du$

2. Expand and simplify the numerator:

   $u(u + 2) = u^2 + 2u$

   So the integral becomes:

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

3. Perform polynomial long division of $\frac{u^2 + 2u}{u - 1}$:

   • Divide $u^2$ by $u$: quotient is $u$.
   • Multiply $u$ by $(u - 1)$: result is $u^2 - u$.
   • Subtract $u^2 - u$ from $u^2 + 2u$: remainder is $3u$.
   • Divide $3u$ by $u - 1$: quotient is $3$, remainder is $3$.

   Thus:

   $\frac{u^2 + 2u}{u - 1} = u + 3 + \frac{3}{u - 1}$

   Substituting this back:

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

4. Break the integral into parts:

   $2 \int u \, du + 2 \int 3 \, du + 2 \int \frac{3}{u - 1} \, du$

   Solve each term:

   • $2 \int u \, du = u^2 + C_1$
   • $2 \int 3 \, du = 6u + C_2$
   • $2 \int \frac{3}{u - 1} \, du = 6 \ln|u - 1| + C_3$

5. Combine the results:

   $u^2 + 6u + 6 \ln|u - 1| + C$

6. Substitute back $u = \sqrt{x}$:

   $\sqrt{x}^2 + 6\sqrt{x} + 6 \ln|\sqrt{x} - 1| + C$

Final Answer:

$x + 6\sqrt{x} + 6 \ln|\sqrt{x} - 1| + C$

Let me know if you'd like further clarification or additional steps! Here are 5 related questions for deeper exploration:

1. How do you handle polynomial division in integrals?
2. What substitution would be useful if $\sqrt{x} - 1$ had a power greater than 1 in the denominator?
3. How does logarithmic integration arise in rational functions?
4. What role does substitution play in simplifying square roots in integrals?
5. Can this integral be solved using partial fractions instead of substitution?

Tip: Always check whether substitution or a direct approach simplifies an integral faster!