https://r2.erweima.ai/mathbot/image/1725759111861oxb6tqvk.jpg

Let's work through the integrals one by one.

### 1. \(\int (1 + 6x)^4(6) \, dx\)

This is a standard substitution problem. Let:
\[
u = 1 + 6x \quad \Rightarrow \quad du = 6 \, dx
\]
Substitute into the integral:
\[
\int u^4 \, du = \frac{u^5}{5} + C
\]
Substitute \(u = 1 + 6x\) back:
\[
\frac{(1 + 6x)^5}{5} + C
\]

---

### 2. \(\int \sqrt{25 - x^2}(-2x) \, dx\)

We can use substitution here as well. Let:
\[
u = 25 - x^2 \quad \Rightarrow \quad du = -2x \, dx
\]
So the integral becomes:
\[
\int \sqrt{u} \, du = \frac{2}{3} u^{3/2} + C
\]
Substitute back \(u = 25 - x^2\):
\[
\frac{2}{3}(25 - x^2)^{3/2} + C
\]

---

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

Let \(u = x^4 + 3\), so:
\[
du = 4x^3 \, dx
\]
The integral becomes:
\[
\frac{1}{4} \int u^2 \, du = \frac{1}{4} \cdot \frac{u^3}{3} + C = \frac{u^3}{12} + C
\]
Substitute back \(u = x^4 + 3\):
\[
\frac{(x^4 + 3)^3}{12} + C
\]

---

### 4. \(\int x^2(x^3 - 1)^4 \, dx\)

Let \(u = x^3 - 1\), so:
\[
du = 3x^2 \, dx
\]
The integral becomes:
\[
\frac{1}{3} \int u^4 \, du =

Solve the integrals provided in the image.

Learn how to solve complex integrals using the substitution method, covering problems involving polynomial and square root expressions. Suitable for students studying calculus at the undergraduate level.

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