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Problem: Solve the given indefinite integrals using integration by parts.

The formula for integration by parts is:
\[
\int u \, dv = uv - \int v \, du
\]
where \( u \) and \( dv \) are chosen such that \( du \) and \( v \) can be easily computed.

Let's solve each problem step by step.

---

1. \(\int e^x \sin(x) \, dx\)

Given:
\[
u = \sin(x), \quad dv = e^x \, dx
\]

#### Step 1: Compute \( du \) and \( v \)
\[
du = \cos(x) \, dx, \quad v = \int e^x \, dx = e^x
\]

#### Step 2: Apply integration by parts
\[
\int e^x \sin(x) \, dx = u v - \int v \, du
\]
\[
= \sin(x) e^x - \int e^x \cos(x) \, dx
\]

#### Step 3: Apply integration by parts again to \(\int e^x \cos(x) \, dx\)
Let:
\[
u_1 = \cos(x), \quad dv_1 = e^x \, dx
\]
\[
du_1 = -\sin(x) \, dx, \quad v_1 = e^x
\]

\[
\int e^x \cos(x) \, dx = u_1 v_1 - \int v_1 \, du_1
\]
\[
= \cos(x) e^x - \int e^x (-\sin(x)) \, dx
\]
\[
= \cos(x) e^x + \int e^x \sin(x) \, dx
\]

#### Step 4: Substitute back into the original equation
\[
\int e^x \sin(x) \, dx = \sin(x) e^x - \left( \cos(x) e^x + \int e^x \sin(x) \, dx \right)
\]
\[
\int e^x \sin(x) \, dx = \sin(x) e^x - \cos(x) e^x - \int e^x \sin(x) \, dx
\]

#### Step 5: Solve for \(\int e^x \sin(x) \, dx\)
Add \(\int e^x \sin(x) \, dx\) to both sides:
\[
2 \int e^x \sin(x) \, dx = \sin(x) e^x - \cos(x) e^x
\]
\[
\int e^x \sin(x) \, dx = \frac{1}{2} \left( \sin(x) e^x - \cos(x) e^x \right) + C
\]

#### Final Answer:
\[
\boxed{\frac{1}{2} e^x (\sin(x) - \cos(x)) + C}
\]

---

2. \(\int x^2 e^{4x} \, dx\)

Given:
\[
u = x^2, \quad dv = e^{4x} \, dx
\]

#### Step 1: Compute \( du \) and \( v \)
\[
du = 2x \, dx, \quad v = \int e^{4x} \, dx = \frac{1}{4} e^{4x}
\]

#### Step 2: Apply integration by parts
\[
\int x^2 e^{4x} \, dx = u v - \int v \, du
\]
\[
= x^2 \cdot \frac{1}{4} e^{4x} - \int \frac{1}{4} e^{4x} \cdot 2x \, dx
\]
\[
= \frac{1}{4} x^2 e^{4x} - \frac{1}{2} \int x e^{4x} \, dx
\]

#### Step 3: Apply integration by parts to \(\int x e^{4x} \, dx\)
Let:
\[
u_1 = x, \quad dv_1 = e^{4x} \, dx
\]
\[
du_1 = dx, \quad v_1 = \frac{1}{4} e^{4x}
\]

\[
\int x e^{4x} \, dx = u_1 v_1 - \int v_1 \, du_1
\]
\[
= x \cdot \frac{1}{4} e^{4x} - \int \frac{1}{4} e^{4x} \, dx
\]
\[
= \frac{1}{4} x e^{4x} - \frac{1}{4} \int e^{4x} \, dx
\]
\[
= \frac{1}{4} x e^{4x} - \frac{1}{4} \cdot \frac{1}{4} e^{4x}
\]
\[
= \frac{1}{4} x e^{4x} - \frac{1}{16} e^{4x}
\]

#### Step 4: Substitute back into the original equation
\[
\int x^2 e^{4x} \, dx = \frac{1}{4} x^2 e^{4x} - \frac{1}{2} \left( \frac{1}{4} x e^{4x} - \frac{1}{16} e^{4x} \right)
\]
\[
= \frac{1}{4} x^2 e^{4x} - \frac{1}{8} x e^{4x} + \frac{1}{32} e^{4x}
\]

#### Final Answer:
\[
\boxed{\frac{1}{4} x^2 e^{4x} - \frac{1}{8} x e^{4x} + \frac{1}{32} e^{4x} + C}
\]

---

3. \(\int \frac{\ln(x)}{x} \, dx\)

Given:
\[
u = \ln(x), \quad dv = \frac{1}{x} \, dx
\]

#### Step 1: Compute \( du \) and \( v \)
\[
du = \frac{1}{x} \, dx, \quad v = \int \frac{1}{x} \, dx = \ln(x)
\]

#### Step 2: Apply integration by parts
\[
\int \frac{\ln(x)}{x} \, dx = u v - \int v \, du
\]
\[
= \ln(x) \cdot \ln(x) - \int \ln(x) \cdot \frac{1}{x} \, dx
\]
\[
= (\ln(x))^2 - \int \frac{\ln(x)}{x} \, dx
\]

#### Step 3: Solve for \(\int \frac{\ln(x)}{x} \, dx\)
Let \( I = \int \frac{\ln(x)}{x} \, dx \). Then:
\[
I = (\ln(x))^2 - I
\]
\[
2I = (\ln(x))^2
\]
\[
I = \frac{1}{2} (\ln(x))^2
\]

#### Final Answer:
\[
\boxed{\frac{1}{2} (\ln(x))^2 + C}
\]

---

4. \(\int x^2 \sin(10x) \, dx\)

Given:
\[
u = x^2, \quad dv = \sin(10x) \, dx
\]

#### Step 1: Compute \( du \) and \( v \)
\[
du = 2x \, dx, \quad v = \int \sin(10x) \, dx = -\frac{1}{10} \cos(10x)
\]

#### Step 2: Apply integration by parts
\[
\int x^2 \sin(10x) \, dx = u v - \int v \, du
\]
\[
= x^2 \left( -\frac{1}{10} \cos(10x) \right) - \int \left( -\frac{1}{10} \cos(10x) \right) (2x) \, dx
\]
\[
= -\frac{1}{10} x^2 \cos(10x) + \frac{1}{5} \int x \cos(10x) \, dx
\]

#### Step 3: Apply integration by parts to \(\int x \cos(10x) \, dx\)
Let:
\[
u_1 = x, \quad dv_1 = \cos(10x) \, dx
\]
\[
du_1 = dx, \quad v_1 = \int \cos(10x) \, dx = \frac{1}{10} \sin(10x)
\]

\[
\int x \cos(10x) \, dx = u_1 v_1 - \int v_1 \, du_1
\]
\[
= x \cdot \frac{1}{10} \sin(10x) - \int \frac{1}{10} \sin(10x) \, dx
\]
\[
= \frac{1}{10} x \sin(10x) - \frac{1}{10} \int \sin(10x) \, dx
\]
\[
= \frac{1}{10} x \sin(10x) - \frac{1}{10} \left( -\frac{1}{10} \cos(10x) \right)
\]
\[
= \frac{1}{10} x \sin(10x) + \frac{1}{100} \cos(10x)
\]

#### Step 4: Substitute back into the original equation
\[
\int x^2 \sin(10x) \, dx = -\frac{1}{10} x^2 \cos(10x) + \frac{1}{5} \left( \frac{1}{10} x \sin(10x) + \frac{1}{100} \cos(10x) \right)
\]
\[
= -\frac{1}{10} x^2 \cos(10x) + \frac{1}{50} x \sin(10x) + \frac{1}{500} \cos(10x)
\]

#### Final Answer:
\[
\boxed{-\frac{1}{10} x^2 \cos(10x) + \frac{1}{50} x \sin(10x) + \frac{1}{500} \cos(10x) + C}
\]

---

5. \(\int x^6 \sqrt{x^3 + 9} \, dx\)

Given:
\[
u = x^3, \quad dv = x^2 \sqrt{x^3 + 9} \, dx
\]

This integral does not directly fit the integration by parts method as stated. Instead, it requires a substitution. Let:
\[
w = x^3 + 9 \implies dw = 3x^2 \, dx \implies x^2 \, dx = \frac{1}{3} dw
\]

Rewrite the integral:
\[
\int x^6 \sqrt{x^3 + 9} \, dx = \int (x^3)^2 \sqrt{x^3 + 9} \cdot x^2 \, dx
\]
\[
= \int (w - 9)^2 \sqrt{w} \cdot \frac{1}{3} \, dw
\]
\[
= \frac{1}{3} \int (w^2 - 18w + 81) w^{1/2} \, dw
\]
\[
= \frac{1}{3} \int \left( w^{5/2} - 18w^{3/2} + 81w^{1/2} \right) \, dw
\]

Integrate term by term:
\[
= \frac{1}{3} \left( \frac{2}{7} w^{7/2} - 18 \cdot \frac{2}{5} w^{5/2} + 81 \cdot \frac{2}{3} w^{3/2} \right) + C
\]
\[
= \frac{1}{3} \left( \frac{2}{7} w^{7/2} - \frac{36}{5} w^{5/2} + 54 w^{3/2} \right) + C
\]
\[
= \frac{2}{21} w^{7/2} - \frac{12}{5} w^{5/2} + 18 w^{3/2} + C
\]

Substitute back \( w = x^3 + 9 \):
\[
= \frac{2}{21} (x^3 + 9)^{7/2} - \frac{12}{5} (x^3 + 9)^{5/2} + 18 (x^3 + 9)^{3/2} + C
\]

#### Final Answer:
\[
\boxed{\frac{2}{21} (x^3 + 9)^{7/2} - \frac{12}{5} (x^3 + 9)^{5/2} + 18 (x^3 + 9)^{3/2} + C}
\]

---

6. \(\int x \sqrt{x + 2} \, dx\)

Given:
\[
u = x, \quad dv = \sqrt{x + 2} \, dx
\]

#### Step 1: Compute \( du \) and \( v \)
\[
du = dx, \quad v = \int \sqrt{x + 2} \, dx = \int (x + 2)^{1/2} \, dx = \frac{2}{3} (x + 2)^{3/2}
\]

#### Step 2: Apply integration by parts
\[
\int x \sqrt{x + 2} \, dx = u v - \int v \, du
\]
\[
= x \cdot \frac{2}{3} (x + 2)^{3/2} - \int \frac{2}{3} (x + 2)^{3/2} \, dx
\]
\[
= \frac{2}{3} x (x + 2)^{3/2} - \frac{2}{3} \int (x + 2)^{3/2} \, dx
\]

#### Step 3: Integrate \(\int (x + 2)^{3/2} \, dx\)
\[
\int (x + 2)^{3/2} \, dx = \frac{2}{5} (x + 2)^{5/2}
\]

#### Step 4: Substitute back
\[
\int x \sqrt{x + 2} \, dx = \frac{2}{3} x (x + 2)^{3/2} - \frac{2}{3} \cdot \frac{2}{5} (x + 2)^{5/2}
\]
\[
= \frac{2}{3} x (x + 2)^{3/2} - \frac{4}{15} (x + 2)^{5/2}
\]

#### Final Answer:
\[
\boxed{\frac{2}{3} x (x + 2)^{3/2} - \frac{4}{15} (x + 2)^{5/2} + C}
\]

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Final Answers:

1. \(\boxed{\frac{1}{2} e^x (\sin(x) - \cos(x)) + C}\)
2. \(\boxed{\frac{1}{4} x^2 e^{4x} - \frac{1}{8} x e^{4x} + \frac{1}{32} e^{4x} + C}\)
3. \(\boxed{\frac{1}{2} (\ln(x))^2 + C}\)
4. \(\boxed{-\frac{1}{10} x^2 \cos(10x) + \frac{1}{50} x \sin(10x) + \frac{1}{500} \cos(10x) + C}\)
5. \(\boxed{\frac{2}{21} (x^3 + 9)^{7/2} - \frac{12}{5} (x^3 + 9)^{5/2} + 18 (x^3 + 9)^{3/2} + C}\)
6. \(\boxed{\frac{2}{3} x (x + 2)^{3/2} - \frac{4}{15} (x + 2)^{5/2} + C}\)