`int (1)/(4e^x+9e^(-x))dx=....`

Answer

To solve the integral \( \int \frac{1}{4e^x + 9e^{-x}} \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
We start with the integral:
\[ I = \int \frac{1}{4e^x + 9e^{-x}} \, dx \]
We can rewrite \( e^{-x} \) as \( \frac{1}{e^x} \):
\[ I = \int \frac{1}{4e^x + \frac{9}{e^x}} \, dx \]

2. **Combine the Terms**:
To combine the terms in the denominator, we can multiply the numerator and denominator by \( e^x \):
\[ I = \int \frac{e^x}{4e^{2x} + 9} \, dx \]

3. **Substitution**:
Let \( t = e^x \). Then, \( dt = e^x \, dx \) or \( dx = \frac{dt}{t} \):
\[ I = \int \frac{1}{4t^2 + 9} \, dt \]

4. **Recognize the Form**:
The integral \( \int \frac{1}{4t^2 + 9} \, dt \) can be recognized as a standard integral of the form \( \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) + C \). Here, \( a^2 = 9 \) implies \( a = 3 \):
\[ I = \frac{1}{3} \tan^{-1} \left( \frac{t}{\frac{3}{2}} \right) + C \]

5. **Substituting Back**:
Since \( t = e^x \), we substitute back:
\[ I = \frac{1}{3} \tan^{-1} \left( \frac{2e^x}{3} \right) + C \]

6. **Final Result**:
Thus, the final result of the integral is:
\[ \int \frac{1}{4e^x + 9e^{-x}} \, dx = \frac{1}{6} \tan^{-1} \left( \frac{2e^x}{3} \right) + C \]