The antiderivative of `(3^(x))/(sqrt(1-9^(x)))` with respect to x is

Options

A. `(log_(3)e)sin^(-1)(3^(x))+C`
B. `sin^(-1)(3^(x))+C`
C. `(log_(3)e)cos^(-1)(3^(x))`
D. none of these

Answer

To find the antiderivative of the function \( \frac{3^x}{\sqrt{1 - 9^x}} \) with respect to \( x \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the integral:
\[ \int \frac{3^x}{\sqrt{1 - 9^x}} \, dx \]
Notice that \( 9^x = (3^2)^x = (3^x)^2 \). Therefore, we can rewrite the expression inside the square root:
\[ 1 - 9^x = 1 - (3^x)^2 \]
Thus, we can express the integral as:
\[ \int \frac{3^x}{\sqrt{1 - (3^x)^2}} \, dx \]

### Step 2: Use substitution
Let us use the substitution:
\[ t = 3^x \]
Then, differentiating both sides gives:
\[ dt = 3^x \ln(3) \, dx \quad \Rightarrow \quad dx = \frac{dt}{3^x \ln(3)} = \frac{dt}{t \ln(3)} \]

### Step 3: Substitute into the integral
Substituting \( t \) into the integral, we have:
\[ \int \frac{t}{\sqrt{1 - t^2}} \cdot \frac{dt}{t \ln(3)} = \frac{1}{\ln(3)} \int \frac{1}{\sqrt{1 - t^2}} \, dt \]

### Step 4: Integrate using a known formula
The integral \( \int \frac{1}{\sqrt{1 - t^2}} \, dt \) is a standard integral that evaluates to:
\[ \sin^{-1}(t) + C \]
Thus, we have:
\[ \frac{1}{\ln(3)} \left( \sin^{-1}(t) + C \right) = \frac{1}{\ln(3)} \sin^{-1}(3^x) + C \]

### Step 5: Final expression
Therefore, the antiderivative of the original function is:
\[ \int \frac{3^x}{\sqrt{1 - 9^x}} \, dx = \frac{1}{\ln(3)} \sin^{-1}(3^x) + C \]

### Conclusion
The final answer can be expressed in logarithmic form:
\[ = \log_3(e) \sin^{-1}(3^x) + C \]