To find the integral of $\frac{\log(x^2 + 1)}{x}$, we proceed as follows:

Given:

$\int \frac{\log(x^2 + 1)}{x} \, dx$

Step 1: Substitution

Let $u = x^2 + 1$, hence: $du = 2x \, dx \quad \text{or} \quad \frac{du}{2} = x \, dx$ Thus, the integral becomes: $\int \frac{\log(u)}{x} \cdot \frac{du}{2x} = \frac{1}{2} \int \frac{\log(u)}{u-1} \, du$

Step 2: Use of the Spence function (also known as the dilogarithm function)

The integral $\int \frac{\log(u)}{u} \, du$ is related to the dilogarithm function, denoted by $\text{Li}_2(u)$.

We proceed with integration by parts: Let $v = \log(u)$ and $dw = \frac{1}{u} du$.

Then: $dv = \frac{1}{u} du \quad \text{and} \quad w = \log(u)$ The integration by parts formula is: $\int v \, dw = vw - \int w \, dv$ So, $\frac{1}{2} \int \log(u) \, d\log(u) = \frac{1}{2} \left[ \log(u)\log(\log(u)) - \int \frac{du}{u}\right]$ Since the integral of $\frac{1}{u}$ is $\log(u)$, we substitute back $u = x^2 + 1$, and obtain the result: $\frac{1}{2} \left[ \text{Li}_2(-x^2) + C\right]$

Thus, the integral simplifies to: $\int \frac{\log(x^2 + 1)}{x} \, dx = \frac{1}{2} \log(x^2 + 1) \log(\log(x^2 + 1)) + \text{Li}_2(-x^2) + C$

Where $\text{Li}_2(u)$ is the dilogarithm function, and $C$ is the constant of integration.

This result involves both logarithmic and special functions (specifically, the dilogarithm function).

Would you like further details or have any questions?

Follow-up Questions:

1. What are the properties of the dilogarithm function $\text{Li}_2(x)$?
2. How can the integral of a logarithmic function be solved using integration by parts?
3. What is the substitution method in calculus, and how is it applied here?
4. How does the dilogarithm function $\text{Li}_2(x)$ relate to other special functions?
5. What are some applications of the dilogarithm function in mathematical analysis?
6. How do you recognize when to apply integration by parts in solving integrals?
7. Can the integral of $\frac{\log(x^2 + 1)}{x}$ be solved without using special functions?
8. How does the natural logarithm function differ from the dilogarithm function?

Tip:

When solving integrals involving logarithmic functions, consider if special functions or substitutions could simplify the process.