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Let's go through the integrals one by one using substitution as requested:

1) \[
\int x^4(1 + x^5)^8 \, dx
\]
   - **Substitute**: Let \( u = 1 + x^5 \), hence \( du = 5x^4 \, dx \).
   - The integral simplifies to:
   \[
   \frac{1}{5} \int u^8 \, du = \frac{1}{5} \cdot \frac{u^9}{9} + C = \frac{(1 + x^5)^9}{45} + C
   \]

2) \[
\int 2x \cos(x^2) \, dx
\]
   - **Substitute**: Let \( u = x^2 \), hence \( du = 2x \, dx \).
   - The integral becomes:
   \[
   \int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C
   \]

3) \[
\int \frac{x}{\sqrt{2 + x^2}} \, dx
\]
   - **Substitute**: Let \( u = 2 + x^2 \), hence \( du = 2x \, dx \).
   - The integral simplifies to:
   \[
   \frac{1}{2} \int \frac{du}{\sqrt{u}} = \sqrt{u} + C = \sqrt{2 + x^2} + C
   \]

4) \[
\int \frac{\ln(x)}{x} \, dx
\]
   - **Substitute**: Let \( u = \ln(x) \), hence \( du = \frac{1}{x} \, dx \).
   - The integral becomes:
   \[
   \int u \, du = \frac{u^2}{2} + C = \frac{\ln^2(x)}{2} + C
   \]

5) \[
\int \frac{\sin(\sqrt{x})}{\sqrt{x}} \, dx
\]
   - **Substitute**: Let \( u = \sqrt{x} \), hence \( du = \frac{1}{2\sqrt{x}} \, dx \).
   - The integral becomes:
   \[
   2 \int \sin(u) \, du = -2 \cos(u) + C = -2 \cos(\sqrt{x}) + C
   \]

6) \[
\int x \cos(4x^2) \, dx
\]
   - **Substitute**: Let \( u = 4x^2 \), hence \( du = 8x \, dx \).
   - The integral becomes:
   \[
   \frac{1}{8} \int \cos(u) \, du = \frac{1}{8} \sin(u) + C = \frac{\sin(4x^2)}{8} + C
   \]

7) \[
\int \frac{1}{x \ln(x)} \, dx
\]
   - **Substitute**: Let \( u = \ln(x) \), hence \( du = \frac{1}{x} \, dx \).
   - The integral becomes:
   \[
   \int \frac{1}{u} \, du = \ln|u| + C = \ln|\ln(x)| + C
   \]

8) \[
\int \tan(x) \, dx
\]
   - Recall that \( \int \tan(x) \, dx = \ln|\sec(x)| + C \).

9) \[
\int x^2 \ln(x) \, dx
\]
   - **Integration by parts**: Let \( u = \ln(x) \), \( dv = x^2 \, dx \), hence \( du = \frac{1}{x} \, dx \), \( v = \frac{x^3}{3} \).
   - The integral becomes:
   \[
   \frac{x^3 \ln(x)}{3} - \int \frac{x^3}{3} \cdot \frac{1}{x} \, dx = \frac{x^3 \ln(x)}{3} - \frac{x^3}{9} + C
   \]

10) \[
\int \frac{e^{\sqrt{x}}}{\sqrt{x}} \, dx
\]
   - **Substitute**: Let \( u = \sqrt{x} \), hence \( du = \frac{1}{2\sqrt{x}} \, dx \).
   - The integral becomes:
   \[
   2 \int e^u \, du = 2e^u + C = 2e^{\sqrt{x}} + C
   \]

Let me know if you'd like further explanations or if anything is unclear!

Here are 5 related questions:
1. How is substitution chosen in integration problems?
2. How do you recognize when to use integration by parts?
3. What is the difference between indefinite and definite integrals?
4. How do you integrate functions involving logarithms?
5. Can you apply the substitution method to all integrals?

**Tip**: Always verify your substitution by differentiating your result to check if you recover the original integrand.

Aplicando el método de sustitución, hallar las siguientes integrales indefinidas.

Learn how to solve indefinite integrals using the substitution method with detailed examples. Problems include trigonometric, logarithmic, and exponential functions. Suitable for students studying calculus.

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