Step 1: Understanding the Concept:
This integral involves exponential functions in the denominator. A standard approach to tackle this is using integration by substitution to transform it into a rational function, which can then be integrated using partial fraction decomposition.
Step 2: Key Formula or Approach:
1. Substitution: Let $t = e^x$, then $dt = e^x dx \implies dx = \frac{dt}{t}$.
2. Partial Fraction Decomposition: Break down $\frac{1}{t(at^2+bt+c)}$ into simpler fractions $\frac{A}{t} + \frac{B}{rt+p} + \frac{C}{st+q}$.
Step 3: Detailed Explanation:
Let the integral be $I = \int \frac{1}{2e^{2x} + 3e^x + 1} dx$.
Substitute $e^x = t$. This gives $e^x dx = dt$, so $dx = \frac{dt}{t}$.
Substituting these into the integral:
\[ I = \int \frac{1}{2t^2 + 3t + 1} \cdot \frac{dt}{t} = \int \frac{dt}{t(2t^2 + 3t + 1)} \]
Factor the quadratic expression in the denominator: $2t^2 + 3t + 1 = 2t^2 + 2t + t + 1 = 2t(t+1) + 1(t+1) = (2t+1)(t+1)$.
\[ I = \int \frac{dt}{t(2t+1)(t+1)} \]
Apply partial fractions:
\[ \frac{1}{t(2t+1)(t+1)} = \frac{A}{t} + \frac{B}{2t+1} + \frac{C}{t+1} \]
Multiply through by the common denominator $t(2t+1)(t+1)$:
\[ 1 = A(2t+1)(t+1) + Bt(t+1) + Ct(2t+1) \]
Solve for the constants $A, B, C$ by substituting strategic values for $t$:
Let $t = 0$: $1 = A(1)(1) \implies A = 1$
Let $t = -1$: $1 = C(-1)(-2+1) \implies 1 = C(-1)(-1) \implies C = 1$
Let $t = -1/2$: $1 = B(-1/2)(-1/2 + 1) \implies 1 = B(-1/2)(1/2) \implies 1 = -B/4 \implies B = -4$
The partial fraction decomposition is:
\[ \frac{1}{t(2t+1)(t+1)} = \frac{1}{t} - \frac{4}{2t+1} + \frac{1}{t+1} \]
Now integrate each term with respect to $t$:
\[ I = \int \left( \frac{1}{t} - \frac{4}{2t+1} + \frac{1}{t+1} \right) dt \]
\[ I = \log|t| - \frac{4}{2}\log|2t+1| + \log|t+1| + c \]
\[ I = \log|t| - 2\log|2t+1| + \log|t+1| + c \]
Substitute back $t = e^x$. Since $e^x$ is always positive, absolute values are unnecessary:
\[ I = \log(e^x) - 2\log(2e^x+1) + \log(e^x+1) + c \]
Since $\log(e^x) = x$, we get the final form:
\[ I = x + \log(e^x+1) - 2\log(2e^x+1) + c \]
Step 4: Final Answer:
The integrated expression is $x + \log (\text{e}^x + 1) - 2 \log (2\text{e}^x + 1) + \text{c}$.