The given integral is:

$\int \frac{x^2 + 1}{x^3 + 3x + 1} \, dx$

To solve this, we can attempt to simplify it by considering a substitution. A common approach in such cases is to use logarithmic differentiation when the degree of the numerator is lower than or equal to that of the denominator (as it is here).

Step 1: Substitution

Let's try the substitution:

$u = x^3 + 3x + 1$ Then, differentiate $u$ with respect to $x$:

$\frac{du}{dx} = 3x^2 + 3$ or equivalently:

$du = (3x^2 + 3) dx$

Now, rewrite $du$ as:

$du = 3(x^2 + 1) \, dx$

So,

$\frac{du}{3} = (x^2 + 1) \, dx$

Step 2: Rewrite the integral

We now rewrite the integral in terms of $u$:

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

This simplifies to:

$\frac{1}{3} \int \frac{du}{u}$

Step 3: Solve the integral

The integral of $\frac{1}{u}$ is:

$\int \frac{du}{u} = \ln|u| + C$

Thus, the solution is:

$\frac{1}{3} \ln|u| + C$

Finally, substitute back $u = x^3 + 3x + 1$ to get the final answer:

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

Final Answer:

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

Would you like further details on this solution?

Here are five related questions to explore this concept further:

1. How do you identify the right substitution in integration problems like this one?
2. What is the significance of the logarithmic function in integrals of rational functions?
3. Can the substitution method always be applied when integrating rational functions?
4. How can you handle cases where the denominator is more complex in form?
5. How do you apply integration by parts in situations where substitution isn't straightforward?

Tip: For rational functions, always check if the degree of the numerator and denominator suggests a possible substitution or logarithmic form.