`int(1)/(3e^(x)+2e^(-x))dx=`

Options

A. `(1)/(sqrt6)tan^(-1)[sqrt((3)/(2))e^(x)]+c`
B. `sqrt6tan^(-1)[(3e^(x))/(2)]+c`
C. `(1)/(sqrt6)tan^(-1)[sqrt((2)/(3))e^(x)]+c`
D. `6tan^(-1)(sqrt6e^(x))+c`

Answer

To solve the integral \( \int \frac{1}{3e^x + 2e^{-x}} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
We start by rewriting the expression in the denominator:

\[ 3e^x + 2e^{-x} = 3e^x + \frac{2}{e^x} \]

Thus, we can rewrite the integral as:

\[ \int \frac{1}{3e^x + 2e^{-x}} \, dx = \int \frac{e^x}{3e^{2x} + 2} \, dx \]

### Step 2: Substitution
Let \( t = e^x \). Then, the differential \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \).

Substituting these into the integral gives:

\[ \int \frac{t}{3t^2 + 2} \cdot \frac{dt}{t} = \int \frac{1}{3t^2 + 2} \, dt \]

### Step 3: Factor Out Constants
Now we can factor out constants from the denominator:

\[ \int \frac{1}{3(t^2 + \frac{2}{3})} \, dt = \frac{1}{3} \int \frac{1}{t^2 + \frac{2}{3}} \, dt \]

### Step 4: Recognize the Integral Form
The integral \( \int \frac{1}{t^2 + a^2} \, dt \) has a known solution:

\[ \int \frac{1}{t^2 + a^2} \, dt = \frac{1}{a} \tan^{-1} \left( \frac{t}{a} \right) + C \]

In our case, \( a^2 = \frac{2}{3} \) implies \( a = \sqrt{\frac{2}{3}} \).

### Step 5: Solve the Integral
Now we can substitute \( a \) back into the integral:

\[ \frac{1}{3} \cdot \frac{1}{\sqrt{\frac{2}{3}}} \tan^{-1} \left( \frac{t}{\sqrt{\frac{2}{3}}} \right) + C \]

This simplifies to:

\[ \frac{1}{3} \cdot \frac{\sqrt{3}}{\sqrt{2}} \tan^{-1} \left( \frac{t \sqrt{3}}{\sqrt{2}} \right) + C \]

### Step 6: Substitute Back for \( t \)
Since \( t = e^x \):

\[ \frac{\sqrt{3}}{3\sqrt{2}} \tan^{-1} \left( \frac{e^x \sqrt{3}}{\sqrt{2}} \right) + C \]

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

\[ \int \frac{1}{3e^x + 2e^{-x}} \, dx = \frac{\sqrt{3}}{3\sqrt{2}} \tan^{-1} \left( \frac{e^x \sqrt{3}}{\sqrt{2}} \right) + C \]