`int(3^(x))/(sqrt(9^(x)-1))dx=`

Options

A. `log_(3)(3^(x)+sqrt(9^(x)-1))+c`
B. `log_(3)(3^(x)-sqrt(9^(x)-1))+c`
C. `log_(9)(3^(x)+sqrt(9^(x)-1))+c`
D. `(1)/(log3)sec^(-1)(3^(x))+c`

Answer

To solve the integral \( \int \frac{3^x}{\sqrt{9^x - 1}} \, dx \), we can follow these steps:

### Step 1: Rewrite the integral
We can express \( 9^x \) in terms of \( 3^x \):
\[ 9^x = (3^2)^x = (3^x)^2 \]
Thus, we can rewrite the integral as:
\[ \int \frac{3^x}{\sqrt{(3^x)^2 - 1}} \, dx \]

### Step 2: Substitution
Let \( t = 3^x \). Then, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 3^x \ln(3) \implies dx = \frac{dt}{3^x \ln(3)} = \frac{dt}{t \ln(3)} \]

### Step 3: Substitute in the integral
Now, substitute \( t \) and \( dx \) into the integral:
\[ \int \frac{t}{\sqrt{t^2 - 1}} \cdot \frac{dt}{t \ln(3)} = \frac{1}{\ln(3)} \int \frac{1}{\sqrt{t^2 - 1}} \, dt \]

### Step 4: Solve the integral
The integral \( \int \frac{1}{\sqrt{t^2 - 1}} \, dt \) is a standard integral, which gives:
\[ \int \frac{1}{\sqrt{t^2 - 1}} \, dt = \ln(t + \sqrt{t^2 - 1}) + C \]
Thus, we have:
\[ \frac{1}{\ln(3)} \left( \ln(t + \sqrt{t^2 - 1}) \right) + C \]

### Step 5: Substitute back for \( t \)
Now substitute back \( t = 3^x \):
\[ \frac{1}{\ln(3)} \left( \ln(3^x + \sqrt{(3^x)^2 - 1}) \right) + C \]

### Step 6: Simplify the expression
Using properties of logarithms, we can express this as:
\[ \frac{1}{\ln(3)} \ln(3^x + \sqrt{9^x - 1}) + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \frac{3^x}{\sqrt{9^x - 1}} \, dx = \frac{1}{\ln(3)} \ln(3^x + \sqrt{9^x - 1}) + C \]