Certainly! Let's find the indefinite integral of the function [tex]\( \frac{1}{\sqrt{1 + e^{2x}}} \)[/tex].
To solve the integral:
[tex]\[ \int \frac{1}{\sqrt{1 + e^{2x}}} \, dx \][/tex]
1. Substitute:
Let [tex]\( u = e^{x} \)[/tex], hence [tex]\( du = e^{x} \, dx \)[/tex], or equivalently [tex]\( dx = \frac{du}{u} \)[/tex].
2. Rewrite the integral:
When [tex]\( x \)[/tex] becomes [tex]\( u \)[/tex]:
[tex]\[ 1 + e^{2x} = 1 + u^2 \][/tex]
Thus, the integral becomes:
[tex]\[ \int \frac{1}{\sqrt{1 + u^2}} \cdot \frac{du}{u} \][/tex]
3. Change the variables:
The resulting integral is:
[tex]\[ \int \frac{du}{u \sqrt{1 + u^2}} \][/tex]
4. Simplify and solve the integral:
Let's simplify this integral using a trigonometric substitution.
Use the substitution [tex]\( u = \sinh(t) \)[/tex], which means [tex]\( du = \cosh(t) \, dt \)[/tex] and we also know [tex]\( 1 + \sinh^2(t) = \cosh^2(t) \)[/tex]:
[tex]\[ \int \frac{\cosh(t) \, dt}{\sinh(t) \cosh(t)} = \int \frac{dt}{\sinh(t)} \][/tex]
5. Integrate:
Now, the integral [tex]\(\int \frac{dt}{\sinh(t)} \)[/tex] can be resolved using hyperbolic identities and properties. This integral results in:
[tex]\[ \log(\coth(t) - \csch(t)) \][/tex]
Re-substitute [tex]\( t = \sinh^{-1}(u) \)[/tex]:
[tex]\[ u = \sinh(t) \][/tex]
This eventually leads to the integral in terms of [tex]\( x \)[/tex]:
6. Combine the results:
Taking all these steps and integrations into account, we get the final expression as a logarithmic function:
[tex]\[ \frac{1}{2} \log\left( \sqrt{e^{2x} + 1} - 1 \right) - \frac{1}{2} \log\left( \sqrt{e^{2x} + 1} + 1 \right) \][/tex]
So, the integral [tex]\(\int \frac{1}{\sqrt{1 + e^{2x}}} \, dx\)[/tex] is:
[tex]\[ \frac{1}{2} \log\left( \sqrt{e^{2x} + 1} - 1 \right) - \frac{1}{2} \log\left( \sqrt{e^{2x} + 1} + 1 \right) + C \][/tex]
Where [tex]\( C \)[/tex] is the constant of integration.
