`int (e^(x)dx)/(e^(2x)+4e^(x)+3)`

Answer

To solve the integral

\[ I = \int \frac{e^x \, dx}{e^{2x} + 4e^x + 3}, \]

we will follow these steps:

### Step 1: Substitute \( t = e^x \)
Let \( t = e^x \). Then, differentiating both sides gives us \( dt = e^x \, dx \), or equivalently, \( dx = \frac{dt}{t} \).

### Step 2: Rewrite the integral
Substituting \( e^x \) with \( t \) in the integral, we have:

\[ I = \int \frac{t \, \frac{dt}{t}}{t^2 + 4t + 3} = \int \frac{dt}{t^2 + 4t + 3}. \]

### Step 3: Factor the denominator
Next, we factor the quadratic expression in the denominator:

\[ t^2 + 4t + 3 = (t + 3)(t + 1). \]

So the integral becomes:

\[ I = \int \frac{dt}{(t + 3)(t + 1)}. \]

### Step 4: Use Partial Fraction Decomposition
We can express the integrand using partial fractions:

\[ \frac{1}{(t + 3)(t + 1)} = \frac{A}{t + 3} + \frac{B}{t + 1}. \]

Multiplying through by the denominator \((t + 3)(t + 1)\) gives:

\[ 1 = A(t + 1) + B(t + 3). \]

Expanding this, we get:

\[ 1 = At + A + Bt + 3B = (A + B)t + (A + 3B). \]

Setting coefficients equal, we have:

1. \( A + B = 0 \)
2. \( A + 3B = 1 \)

From the first equation, \( A = -B \). Substituting into the second equation:

\[ -B + 3B = 1 \implies 2B = 1 \implies B = \frac{1}{2}. \]

Thus, \( A = -\frac{1}{2} \).

### Step 5: Rewrite the integral
Now we can rewrite the integral:

\[ I = \int \left( \frac{-\frac{1}{2}}{t + 3} + \frac{\frac{1}{2}}{t + 1} \right) dt. \]

This simplifies to:

\[ I = -\frac{1}{2} \int \frac{dt}{t + 3} + \frac{1}{2} \int \frac{dt}{t + 1}. \]

### Step 6: Integrate
Now we integrate each term:

\[ I = -\frac{1}{2} \ln |t + 3| + \frac{1}{2} \ln |t + 1| + C, \]

where \( C \) is the constant of integration.

### Step 7: Combine the logarithms
Using the properties of logarithms, we can combine these:

\[ I = \frac{1}{2} \left( \ln |t + 1| - \ln |t + 3| \right) + C = \frac{1}{2} \ln \left| \frac{t + 1}{t + 3} \right| + C. \]

### Step 8: Substitute back for \( t \)
Finally, substituting back \( t = e^x \):

\[ I = \frac{1}{2} \ln \left| \frac{e^x + 1}{e^x + 3} \right| + C. \]

### Final Answer
Thus, the final answer is:

\[ I = \frac{1}{2} \ln \left| \frac{e^x + 1}{e^x + 3} \right| + C. \]