To solve the integral [tex]\(\int \frac{e^x \, dx}{e^{2x} + 2e^x + 5}\)[/tex], we can follow these steps:
1. Substitution:
Let [tex]\( u = e^x \)[/tex]. Consequently, [tex]\( du = e^x dx \)[/tex], or [tex]\( dx = \frac{du}{u} \)[/tex].
Thus, the integral becomes:
[tex]\[
\int \frac{u \, \frac{du}{u}}{u^2 + 2u + 5}
\][/tex]
which simplifies to:
[tex]\[
\int \frac{du}{u^2 + 2u + 5}
\][/tex]
2. Completing the square for the denominator:
The quadratic [tex]\( u^2 + 2u + 5 \)[/tex] can be written as:
[tex]\[
u^2 + 2u + 5 = (u+1)^2 + 4
\][/tex]
3. Rewriting the integral:
Now the integral becomes:
[tex]\[
\int \frac{du}{(u+1)^2 + 4}
\][/tex]
4. Substitution for trigonometric integral:
Let [tex]\( v = u + 1 \)[/tex]. Then [tex]\( dv = du \)[/tex], and the integral becomes:
[tex]\[
\int \frac{dv}{v^2 + 4}
\][/tex]
5. Evaluating the integral:
Recognize that the integral [tex]\(\int \frac{dv}{v^2 + a^2} = \frac{1}{a} \arctan \left( \frac{v}{a} \right)\)[/tex] for [tex]\(a = 2\)[/tex]. Therefore, we have:
[tex]\[
\int \frac{dv}{v^2 + 4} = \frac{1}{2} \arctan \left( \frac{v}{2} \right)
\][/tex]
6. Back-substitution:
Recall that [tex]\( v = u + 1 \)[/tex] and [tex]\( u = e^x \)[/tex]:
[tex]\[
\frac{1}{2} \arctan \left( \frac{u + 1}{2} \right) = \frac{1}{2} \arctan \left( \frac{e^x + 1}{2} \right)
\][/tex]
Therefore, the definite integral [tex]\(\int \frac{e^x \, dx}{e^{2x} + 2e^x + 5}\)[/tex] evaluates to:
[tex]\[
\frac{1}{2} \arctan \left( \frac{e^x + 1}{2} \right) + C
\][/tex]
