The integral shown is:

$\int \frac{1 + \sec^3 x}{\cot x} \, dx$

To solve this integral, we can first simplify it by expressing $\cot x$ in terms of $\sin x$ and $\cos x$, and then use substitution techniques if necessary.

Step 1: Express $\cot x$ in terms of sine and cosine.

$\cot x = \frac{\cos x}{\sin x}$

Now the integral becomes: $\int \frac{(1 + \sec^3 x)}{\frac{\cos x}{\sin x}} \, dx = \int \frac{(1 + \sec^3 x) \sin x}{\cos x} \, dx$

Step 2: Express $\sec^3 x$ in terms of $\cos x$.

Recall that $\sec x = \frac{1}{\cos x}$, so: $\sec^3 x = \frac{1}{\cos^3 x}$

Substitute this into the integral: $\int \frac{(1 + \frac{1}{\cos^3 x}) \sin x}{\cos x} \, dx = \int \left( \frac{\sin x}{\cos x} + \frac{\sin x}{\cos^4 x} \right) dx$

Step 3: Split the integral.

Now we have two simpler integrals to deal with: $\int \frac{\sin x}{\cos x} \, dx + \int \frac{\sin x}{\cos^4 x} \, dx$

First integral:

$\int \frac{\sin x}{\cos x} \, dx = \int \tan x \, dx = \ln |\sec x| + C_1$

Second integral:

To handle the second integral, we can use the substitution $u = \cos x$, so $du = -\sin x \, dx$, transforming the integral into: $-\int \frac{1}{u^4} \, du = \frac{1}{3u^3} + C_2 = \frac{1}{3\cos^3 x} + C_2$

Step 4: Combine the results.

Now combine the results from both integrals: $\ln |\sec x| - \frac{1}{3\cos^3 x} + C$

Thus, the solution to the integral is:

$\ln |\sec x| - \frac{1}{3\cos^3 x} + C$

Would you like any further clarification or details on any part of this solution?

Here are 5 related questions to explore further:

1. How do you approach integrals involving trigonometric identities?
2. What are common trigonometric substitutions in integrals?
3. How can we integrate $\sec^n x$ for different powers of $n$?
4. What is the role of logarithmic functions in trigonometric integrals?
5. Can you derive the integral of $\tan x$ without using substitution?

Tip: When solving integrals involving trigonometric functions, it's helpful to recall basic identities and be comfortable using substitutions for simplification.