To solve the integral [tex]\(\int \frac{dx}{e^x + e^{-x}},\)[/tex] we need to strategically evaluate the integrand. Here is a step-by-step solution:
1. Integrand Simplification:
We start by recognizing the expression inside the integral:
[tex]\[
\int \frac{dx}{e^x + e^{-x}}.
\][/tex]
Recall that [tex]\(e^x + e^{-x} = 2\cosh(x)\)[/tex], where [tex]\(\cosh(x)\)[/tex] is the hyperbolic cosine function. Thus, we can rewrite the integral as:
[tex]\[
\int \frac{dx}{2\cosh(x)}.
\][/tex]
2. Using the property of hyperbolic functions:
We know that [tex]\(\cosh(x)\)[/tex] has the property:
[tex]\[
\cosh(x) = \frac{e^x + e^{-x}}{2}.
\][/tex]
Substituting this back into our integral, we get:
[tex]\[
\int \frac{dx}{2\cosh(x)} = \int \frac{dx}{e^x + e^{-x}}.
\][/tex]
Therefore, our integral simplifies to:
[tex]\[
\int \frac{dx}{2\cosh(x)} = \frac{1}{2} \int \frac{dx}{\cosh(x)}.
\][/tex]
3. Integral of [tex]\(\frac{1}{\cosh(x)}\)[/tex]:
The integral of [tex]\(\frac{1}{\cosh(x)}\)[/tex] is a standard result related to the hyperbolic trigonometric function [tex]\(\sech(x)\)[/tex]:
[tex]\[
\int \sech(x) \, dx = 2 \arctan(e^x).
\][/tex]
4. Putting it all together:
We apply the result from the standard integral:
[tex]\[
\int \frac{dx}{e^x + e^{-x}} = \frac{1}{2} \int \sech(x) \, dx = \frac{1}{2} \cdot 2 \arctan(e^x).
\][/tex]
Simplifying this further:
[tex]\[
\int \frac{dx}{e^x + e^{-x}} = \arctan(e^x).
\][/tex]
Therefore, the final result of the integral is:
[tex]\[
\boxed{\arctan(e^x) + C.}
\][/tex]
Note that this solution may not immediately seem to correspond with the given answer, [tex]\(RootSum(4_z^2 + 1, \Lambda(_i, _i\log(-2_i + exp(-x))))\)[/tex], but it follows from standard techniques used in calculus. Both forms of the solution are valid interpretations of the integral’s result, typically converging when simplifying complex expressions.
