To solve the integral \(\int \text{sech}(\ln(x)) \, dx\), we can use a substitution method.

Let \( u = \ln(x) \). Then, \( du = \frac{1}{x} dx \) or equivalently \( dx = x \, du = e^u \, du \).

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

\[
\int \text{sech}(\ln(x)) \, dx = \int \text{sech}(u) \cdot e^u \, du
\]

We recognize that \(\text{sech}(u) = \frac{2}{e^u + e^{-u}}\), so the integral becomes:

\[
\int \frac{2 e^u}{e^u + e^{-u}} \, du
\]

Simplify the integrand:

\[
\int \frac{2 e^u}{e^u + \frac{1}{e^u}} \, du = \int \frac{2 e^u}{e^{2u} + 1} \, du
\]

Notice that \(\frac{2 e^u}{e^{2u} + 1}\) can be integrated directly as it matches the derivative of the arctangent function:

\[
\int \frac{2 e^u}{e^{2u} + 1} \, du = 2 \int \frac{1}{1 + e^{-2u}} \, du
\]

This integral can be recognized as:

\[
2 \int \frac{1}{1 + e^{-2u}} \, du = 2 \arctan(e^u) + C
\]

Now, substituting back \( u = \ln(x) \):

\[
2 \arctan(e^{\ln(x)}) + C = 2 \arctan(x) + C
\]

Thus, the solution to the integral \(\int \text{sech}(\ln(x)) \, dx\) is:

\[
\boxed{2 \arctan(x) + C}
\]

Would you like more details or have any questions?

Here are five more questions you might consider:
1. How do you solve \(\int \frac{1}{x \ln(x)} \, dx\)?
2. What is the integral of \( e^{x^2} \, dx \)?
3. Can you explain the method of partial fractions?
4. How do you find the derivative of \( \tan^{-1}(x) \)?
5. What is the integral of \( \sin^2(x) \, dx \)?

**Tip:** When dealing with integrals involving logarithmic or exponential functions, substitution is often a helpful method to simplify the integral.

Discover how to solve the integral of sech(ln(x)) using the substitution method. This detailed guide covers key concepts like hyperbolic functions, integral substitution, and arctangent derivatives. Ideal for advanced college-level students and anyone interested in calculus.

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