This worksheet provides excellent practice for students learning to integrate logarithmic and exponential functions using fundamental calculus rules.

Problem: Solve the given indefinite integrals using logarithmic and exponential rules.

We will solve each integral step by step. Recall the following key integration formulas:
1. Exponential functions:
\[
\int e^x \, dx = e^x + C
\]
\[
\int a^x \, dx = \frac{a^x}{\ln(a)} + C \quad \text{(for } a > 0 \text{ and } a \neq 1\text{)}
\]

2. Power rule for \( x^n \):
\[
\int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad \text{(for } n \neq -1\text{)}
\]
For \( n = -1 \):
\[
\int x^{-1} \, dx = \int \frac{1}{x} \, dx = \ln|x| + C
\]

3. Constant multiples:
\[
\int c \cdot f(x) \, dx = c \cdot \int f(x) \, dx
\]

Now, let's solve each integral:

---

#### 1. \( \int (9x^{-1}) \, dx \)
- Factor out the constant \( 9 \):
\[
\int (9x^{-1}) \, dx = 9 \int x^{-1} \, dx
\]
- Use the formula for \( \int x^{-1} \, dx \):
\[
\int x^{-1} \, dx = \ln|x|
\]
- Therefore:
\[
9 \int x^{-1} \, dx = 9 \ln|x| + C
\]

Answer:
\[
\boxed{9 \ln|x| + C}
\]

---

#### 2. \( \int \left(\frac{7}{x}\right) \, dx \)
- Rewrite \( \frac{7}{x} \) as \( 7x^{-1} \):
\[
\int \left(\frac{7}{x}\right) \, dx = \int 7x^{-1} \, dx
\]
- Factor out the constant \( 7 \):
\[
\int 7x^{-1} \, dx = 7 \int x^{-1} \, dx
\]
- Use the formula for \( \int x^{-1} \, dx \):
\[
\int x^{-1} \, dx = \ln|x|
\]
- Therefore:
\[
7 \int x^{-1} \, dx = 7 \ln|x| + C
\]

Answer:
\[
\boxed{7 \ln|x| + C}
\]

---

#### 3. \( \int (12e^x) \, dx \)
- Factor out the constant \( 12 \):
\[
\int (12e^x) \, dx = 12 \int e^x \, dx
\]
- Use the formula for \( \int e^x \, dx \):
\[
\int e^x \, dx = e^x
\]
- Therefore:
\[
12 \int e^x \, dx = 12e^x + C
\]

Answer:
\[
\boxed{12e^x + C}
\]

---

#### 4. \( \int (4 \cdot 16^x) \, dx \)
- Factor out the constant \( 4 \):
\[
\int (4 \cdot 16^x) \, dx = 4 \int 16^x \, dx
\]
- Use the formula for \( \int a^x \, dx \):
\[
\int 16^x \, dx = \frac{16^x}{\ln(16)} + C
\]
- Therefore:
\[
4 \int 16^x \, dx = 4 \cdot \frac{16^x}{\ln(16)} + C = \frac{4 \cdot 16^x}{\ln(16)} + C
\]

Answer:
\[
\boxed{\frac{4 \cdot 16^x}{\ln(16)} + C}
\]

---

#### 5. \( \int \left(\frac{8}{x}\right) \, dx \)
- Rewrite \( \frac{8}{x} \) as \( 8x^{-1} \):
\[
\int \left(\frac{8}{x}\right) \, dx = \int 8x^{-1} \, dx
\]
- Factor out the constant \( 8 \):
\[
\int 8x^{-1} \, dx = 8 \int x^{-1} \, dx
\]
- Use the formula for \( \int x^{-1} \, dx \):
\[
\int x^{-1} \, dx = \ln|x|
\]
- Therefore:
\[
8 \int x^{-1} \, dx = 8 \ln|x| + C
\]

Answer:
\[
\boxed{8 \ln|x| + C}
\]

---

#### 6. \( \int (5e^x) \, dx \)
- Factor out the constant \( 5 \):
\[
\int (5e^x) \, dx = 5 \int e^x \, dx
\]
- Use the formula for \( \int e^x \, dx \):
\[
\int e^x \, dx = e^x
\]
- Therefore:
\[
5 \int e^x \, dx = 5e^x + C
\]

Answer:
\[
\boxed{5e^x + C}
\]

---

#### 7. \( \int (7 \cdot 3^x) \, dx \)
- Factor out the constant \( 7 \):
\[
\int (7 \cdot 3^x) \, dx = 7 \int 3^x \, dx
\]
- Use the formula for \( \int a^x \, dx \):
\[
\int 3^x \, dx = \frac{3^x}{\ln(3)} + C
\]
- Therefore:
\[
7 \int 3^x \, dx = 7 \cdot \frac{3^x}{\ln(3)} + C = \frac{7 \cdot 3^x}{\ln(3)} + C
\]

Answer:
\[
\boxed{\frac{7 \cdot 3^x}{\ln(3)} + C}
\]

---

#### 8. \( \int (5x^{-1}) \, dx \)
- Factor out the constant \( 5 \):
\[
\int (5x^{-1}) \, dx = 5 \int x^{-1} \, dx
\]
- Use the formula for \( \int x^{-1} \, dx \):
\[
\int x^{-1} \, dx = \ln|x|
\]
- Therefore:
\[
5 \int x^{-1} \, dx = 5 \ln|x| + C
\]

Answer:
\[
\boxed{5 \ln|x| + C}
\]

---

#### 9. \( \int (-8e^x) \, dx \)
- Factor out the constant \( -8 \):
\[
\int (-8e^x) \, dx = -8 \int e^x \, dx
\]
- Use the formula for \( \int e^x \, dx \):
\[
\int e^x \, dx = e^x
\]
- Therefore:
\[
-8 \int e^x \, dx = -8e^x + C
\]

Answer:
\[
\boxed{-8e^x + C}
\]

---

#### 10. \( \int (-9 \cdot 13^x) \, dx \)
- Factor out the constant \( -9 \):
\[
\int (-9 \cdot 13^x) \, dx = -9 \int 13^x \, dx
\]
- Use the formula for \( \int a^x \, dx \):
\[
\int 13^x \, dx = \frac{13^x}{\ln(13)} + C
\]
- Therefore:
\[
-9 \int 13^x \, dx = -9 \cdot \frac{13^x}{\ln(13)} + C = \frac{-9 \cdot 13^x}{\ln(13)} + C
\]

Answer:
\[
\boxed{\frac{-9 \cdot 13^x}{\ln(13)} + C}
\]

---

Final Answers:

1. \( \boxed{9 \ln|x| + C} \)
2. \( \boxed{7 \ln|x| + C} \)
3. \( \boxed{12e^x + C} \)
4. \( \boxed{\frac{4 \cdot 16^x}{\ln(16)} + C} \)
5. \( \boxed{8 \ln|x| + C} \)
6. \( \boxed{5e^x + C} \)
7. \( \boxed{\frac{7 \cdot 3^x}{\ln(3)} + C} \)
8. \( \boxed{5 \ln|x| + C} \)
9. \( \boxed{-8e^x + C} \)
10. \( \boxed{\frac{-9 \cdot 13^x}{\ln(13)} + C} \)