The text extraction wasn't very clear. However, based on the visible parts, it seems we need to integrate the following functions:

1. $\csc(4x+3)\cot(4x+3)$
2. $\frac{1}{(2x+3)^3}$
3. $\frac{1}{3x+2}$
4. $\csc^2(7-11x)$
5. $e^{6-4x}$

Let's find the integrals for each of these functions:

a) $\csc(4x+3)\cot(4x+3)$

The integral of $\csc(u)\cot(u)$ is $-\csc(u)$. Let $u = 4x + 3$.

$\int \csc(4x+3)\cot(4x+3) \, dx = -\csc(4x+3) \cdot \frac{1}{4} + C = -\frac{1}{4}\csc(4x+3) + C$

b) $\frac{1}{(2x+3)^3}$

Use the substitution $u = 2x + 3$, hence $du = 2dx$.

$\int \frac{1}{(2x+3)^3} \, dx = \int \frac{1}{u^3} \cdot \frac{du}{2} = \frac{1}{2} \int u^{-3} \, du = \frac{1}{2} \cdot \frac{u^{-2}}{-2} + C = -\frac{1}{4(2x+3)^2} + C$

c) $\frac{1}{3x+2}$

Use the substitution $u = 3x + 2$, hence $du = 3dx$.

$\int \frac{1}{3x+2} \, dx = \int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{1}{u} \, du = \frac{1}{3} \ln|u| + C = \frac{1}{3} \ln|3x+2| + C$

d) $\csc^2(7-11x)$

The integral of $\csc^2(u)$ is $-\cot(u)$. Let $u = 7 - 11x$.

$\int \csc^2(7-11x) \, dx = -\cot(7-11x) \cdot \frac{1}{-11} + C = \frac{1}{11}\cot(7-11x) + C$

e) $e^{6-4x}$

Use the substitution $u = 6 - 4x$, hence $du = -4dx$.

$\int e^{6-4x} \, dx = \int e^u \cdot \frac{du}{-4} = -\frac{1}{4} e^u + C = -\frac{1}{4} e^{6-4x} + C$

Would you like details or have any questions on any of these integrals?

Follow-up Questions:

1. Can you provide more examples of integrating trigonometric functions?
2. How do you integrate functions involving square roots?
3. What is the method to integrate rational functions?
4. Can you show how to use partial fraction decomposition for integration?
5. What is the process of integrating by parts?
6. How do you handle integrals involving exponential functions?
7. Can you explain how to integrate using trigonometric substitution?
8. What are improper integrals and how are they evaluated?

Tip:

When integrating complex functions, consider using substitution to simplify the integral into a more recognizable form.