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Here are the solutions to the integrals:

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### **1.** \(\int x^2 \cdot (x^3 + 1)^4 \, dx\)

Let \(u = x^3 + 1\). Then:
- \(du = 3x^2 dx\)
- \(x^2 dx = \frac{1}{3} du\)

Substitute:
\[
\int x^2 \cdot (x^3 + 1)^4 \, dx = \int (u)^4 \cdot \frac{1}{3} du = \frac{1}{3} \int u^4 \, du
\]

Evaluate the integral:
\[
\frac{1}{3} \cdot \frac{u^5}{5} + C = \frac{u^5}{15} + C
\]

Substitute back \(u = x^3 + 1\):
\[
\frac{(x^3 + 1)^5}{15} + C
\]

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### **2.** \(\int e^{2x} \, dx\)

The integral of \(e^{kx}\) is \(\frac{e^{kx}}{k}\). Here \(k = 2\):
\[
\int e^{2x} \, dx = \frac{e^{2x}}{2} + C
\]

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### **3.** \(\int \frac{\sin(3x)}{\cos^2(3x)} \, dx\)

Let \(u = \cos(3x)\). Then:
- \(du = -3\sin(3x) dx\)
- \(-\frac{1}{3} du = \sin(3x) dx\)

Substitute:
\[
\int \frac{\sin(3x)}{\cos^2(3x)} \, dx = \int \frac{-\frac{1}{3} du}{u^2} = -\frac{1}{3} \int u^{-2} \, du
\]

Evaluate the integral:
\[
-\frac{1}{3} \cdot \left(-\frac{1}{u}\right) + C = \frac{1}{3u} + C
\]

Substitute back \(u = \cos(3x)\):
\[
\frac{1}{3\cos(3x)} + C
\]

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### Final Results:
1. \(\frac{(x^3 + 1)^5}{15} + C\)
2. \(\frac{e^{2x}}{2} + C\)
3. \(\frac{1}{3\cos(3x)} + C\)

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Do you want detailed steps for any specific part?  

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#### Related Questions:
1. How would the substitution method change if the powers in problem 1 were different?
2. How does the base of the exponential function affect its integral (like in problem 2)?
3. Can we use trigonometric identities to simplify problem 3 further?
4. What happens when dealing with definite integrals for these problems?
5. How would the solution of problem 3 change for a denominator of \(\cos^3(3x)\)?

#### **Tip:** Always check for substitution opportunities in integrals involving products or compositions of functions.

Solve the integrals provided in the image: (1) ∫x²(x³ + 1)^4 dx, (2) ∫e^(2x) dx, (3) ∫(sin(3x)/cos²(3x)) dx.

Detailed solutions for integrals involving algebraic, exponential, and trigonometric functions, including substitution techniques and key integration formulas. Perfect for calculus students.

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