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Let's solve each integral one by one. We will integrate the given functions with respect to \( x \).

### a) \(\int \frac{e^{x}(1 + x)}{\cos^2(x^2)} \, dx\)

To solve this integral, let's use substitution.

Let \( u = x^2 \), then \( du = 2x \, dx \), hence \( dx = \frac{du}{2x} \).

However, \( \frac{1 + x}{\cos^2(x^2)} \) doesn't lend itself to an immediate substitution that simplifies easily. Thus, we need to try another approach or recognize the function as it stands. This integral is quite complex and does not simplify easily. For exact solutions, numerical methods or specialized functions might be necessary. 

### b) \(\int \frac{4x^3 + 2x}{1 + x^4} \, dx\)

For this integral, we recognize that the derivative of \( x^4 \) is \( 4x^3 \). 

Let \( u = x^4 + 1 \), then \( du = 4x^3 \, dx \), thus \( dx = \frac{du}{4x^3} \).

Now, rewrite the integral in terms of \( u \):

\[
\int \frac{4x^3 + 2x}{1 + x^4} \, dx = \int \frac{4x^3}{1 + x^4} \, dx + \int \frac{2x}{1 + x^4} \, dx
\]

Substituting \( u = 1 + x^4 \):

\[
\int \frac{4x^3}{u} \cdot \frac{du}{4x^3} = \int \frac{1}{u} \, du = \ln|u| = \ln|1 + x^4|
\]

For the second part, we use the substitution \( v = x^2 \):

\[
\int \frac{2x}{1 + x^4} \, dx = \int \frac{2x}{1 + (x^2)^2} \, dx = \int \frac{1}{1 + v^2} \, dv = \arctan(v) = \arctan(x^2)
\]

Therefore, the integral is:

\[
\ln|1 + x^4| + \arctan(x^2) + C
\]

### c) \(\int \frac{1}{x + x \log x} \, dx\)

Let's factor out \( x \):

\[
\int \frac{1}{x (1 + \log x)} \, dx
\]

Let \( u = 1 + \log x \), then \( du = \frac{1}{x} \, dx \), hence \( dx = x \, du \).

So, the integral becomes:

\[
\int \frac{1}{x u} \cdot x \, du = \int \frac{1}{u} \, du = \ln|u| = \ln|1 + \log x| + C
\]

### d) \(\int \tan^3 x \sec x \, dx\)

Rewrite \( \tan^3 x \sec x \) in terms of \( \sec x \) and \( \tan x \):

\[
\tan^3 x \sec x = \tan^2 x \cdot \tan x \sec x = (\sec^2 x - 1) \tan x \sec x
\]

Now, split the integral:

\[
\int (\sec^2 x - 1) \tan x \sec x \, dx = \int \sec^2 x \tan x \, dx - \int \tan x \sec x \, dx
\]

The first part:

\[
\int \sec^2 x \tan x \, dx
\]

Let \( u = \tan x \), then \( du = \sec^2 x \, dx \), thus:

\[
\int u \, du = \frac{u^2}{2} = \frac{\tan^2 x}{2}
\]

The second part:

\[
\int \tan x \sec x \, dx = \int \frac{\sin x}{\cos^2 x} \, dx = -\int \frac{d}{dx}(\frac{1}{\cos x}) \, dx = -\sec x
\]

Combining both parts, we get:

\[
\frac{\tan^2 x}{2} - \sec x + C
\]

### Summary

- a) is quite complex and might require numerical or specialized methods.
- b) \( \ln|1 + x^4| + \arctan(x^2) + C \)
- c) \( \ln|1 + \log x| + C \)
- d) \( \frac{\tan^2 x}{2} - \sec x + C \)

Do you want any more details or have any questions?

### Follow-Up Questions

1. What is the general method for solving integrals involving trigonometric functions?
2. How can substitution be used effectively in integration?
3. What are the common strategies for integrating rational functions?
4. How do you handle integrals that involve logarithmic functions?
5. What numerical methods can be used for complex integrals that do not simplify easily?
6. How can integration by parts be used in solving integrals?
7. What are the different techniques for integrating functions involving polynomials?
8. How do trigonometric identities help in solving integrals?

### Tip

Always look for substitutions that simplify the integral into a standard form or a form that is easier to integrate.

Learn how to solve complex integrals including \( \int \frac{e^{x}(1 + x)}{\cos^2(x^2)} \, dx \), \( \int \frac{4x^3 + 2x}{1 + x^4} \, dx \), \( \int \frac{1}{x + x \log x} \, dx \), and \( \int \tan^3 x \sec x \, dx \). Understand the use of substitution, trigonometric identities, and logarithmic functions in integration. Suitable for students and professionals studying advanced mathematics.

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