`int(dx)/(e^(x)+1-2e^(-x))=`

Options

A. `logabs(e^(x)-1)-logabs(e^(x)+2)+c`
B. `1/2logabs(e^(x)-1)-1/3logabs(e^(x)+2)+c`
C. `1/3logabs(e^(x)-1)-1/3logabs(e^(x)+2)+c`
D. `1/3logabs(e^(x)-1)+1/3logabs(e^(x)+2)+c`

Answer

To solve the integral

\[ I = \int \frac{dx}{e^x + 1 - 2e^{-x}}, \]

we can follow these steps:

### Step 1: Simplify the Denominator
First, we rewrite the denominator:

\[ e^x + 1 - 2e^{-x} = e^x + 1 - \frac{2}{e^x} = \frac{e^{2x} + e^x - 2}{e^x}. \]

Thus, we can rewrite the integral as:

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

### Step 2: Substitute \( t = e^x \)
Let \( t = e^x \). Then, \( dx = \frac{dt}{t} \). Substituting this into the integral gives:

\[ I = \int \frac{t \cdot \frac{dt}{t}}{t^2 + t - 2} = \int \frac{dt}{t^2 + t - 2}. \]

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

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

Thus, we can rewrite the integral as:

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

### Step 4: Partial Fraction Decomposition
Now, we perform partial fraction decomposition:

\[ \frac{1}{(t + 2)(t - 1)} = \frac{A}{t + 2} + \frac{B}{t - 1}. \]

Multiplying through by the denominator \((t + 2)(t - 1)\) gives:

\[ 1 = A(t - 1) + B(t + 2). \]

Expanding and rearranging gives:

\[ 1 = (A + B)t + (2B - A). \]

Setting up the system of equations:

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

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

\[ 2(-A) - A = 1 \implies -3A = 1 \implies A = -\frac{1}{3}, \quad B = \frac{1}{3}. \]

Thus, we have:

\[ \frac{1}{(t + 2)(t - 1)} = -\frac{1}{3(t + 2)} + \frac{1}{3(t - 1)}. \]

### Step 5: Integrate
Now we can integrate:

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

### Step 6: Substitute Back
Substituting back \( t = e^x \):

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

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

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