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 expression \(\frac{3^x}{\sqrt{1 - 9^x}} \, dx\), we can follow these steps:

### Step 1: Rewrite the expression
First, we can rewrite \(9^x\) as \((3^2)^x = (3^x)^2\). Thus, we have:
\[ \frac{3^x}{\sqrt{1 - 9^x}} = \frac{3^x}{\sqrt{1 - (3^x)^2}} \]

### 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, we substitute \(t\) and \(dx\) into the integral:
\[ \int \frac{3^x}{\sqrt{1 - 9^x}} \, dx = \int \frac{t}{\sqrt{1 - t^2}} \cdot \frac{dt}{t \ln(3)} = \frac{1}{\ln(3)} \int \frac{1}{\sqrt{1 - t^2}} \, dt \]

### Step 4: Integrate
The integral \(\int \frac{1}{\sqrt{1 - t^2}} \, dt\) is a standard integral that equals \(\sin^{-1}(t) + C\). Therefore, we have:
\[ \frac{1}{\ln(3)} \int \frac{1}{\sqrt{1 - t^2}} \, dt = \frac{1}{\ln(3)} \sin^{-1}(t) + C \]

### Step 5: Substitute back for \(t\)
Now, we substitute back \(t = 3^x\):
\[ \frac{1}{\ln(3)} \sin^{-1}(3^x) + C \]

### Final Answer
Thus, the antiderivative of \(\frac{3^x}{\sqrt{1 - 9^x}} \, dx\) is:
\[ \frac{1}{\ln(3)} \sin^{-1}(3^x) + C \]