`int(dx)/sqrt(2-3x-x^(2))=fog(x)+C` then

Options

A. `f(x) = sin^(-1)x,g(x)=(2x-3)/(sqrt(17))`
B. `f(x)=tan^(-1) x, g(x) =(2x-3)/(sqrt17)`
C. `f(x)=sin^(-1)x,g(x) = (2x+3)/(17)`
D. none of these

Answer

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

### Step 1: Rewrite the expression under the square root
We start with the expression under the square root:
\[ 2 - 3x - x^2 \]
We can rearrange it as:
\[ -(x^2 + 3x - 2) \]
Next, we will complete the square for the quadratic expression \( x^2 + 3x - 2 \).

### Step 2: Complete the square
To complete the square for \( x^2 + 3x - 2 \):
1. Take half of the coefficient of \( x \) (which is \( 3 \)), square it, and add/subtract it:
\[ \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]
2. Rewrite the expression:
\[ x^2 + 3x + \frac{9}{4} - \frac{9}{4} - 2 = \left(x + \frac{3}{2}\right)^2 - \frac{17}{4} \]
Thus, we can write:
\[ 2 - 3x - x^2 = -\left(\left(x + \frac{3}{2}\right)^2 - \frac{17}{4}\right) = \frac{17}{4} - \left(x + \frac{3}{2}\right)^2 \]

### Step 3: Substitute into the integral
Now we substitute this back into the integral:
\[ I = \int \frac{dx}{\sqrt{\frac{17}{4} - \left(x + \frac{3}{2}\right)^2}} \]
We can factor out \( \frac{17}{4} \) from the square root:
\[ I = \int \frac{dx}{\sqrt{\frac{17}{4}} \sqrt{1 - \frac{4}{17}\left(x + \frac{3}{2}\right)^2}} \]
This simplifies to:
\[ I = \frac{2}{\sqrt{17}} \int \frac{dx}{\sqrt{1 - \frac{4}{17}\left(x + \frac{3}{2}\right)^2}} \]

### Step 4: Use the standard integral formula
The integral \( \int \frac{dx}{\sqrt{1 - k^2 x^2}} \) has the result \( \sin^{-1}(kx) + C \). In our case, we have:
\[ k^2 = \frac{4}{17} \implies k = \frac{2}{\sqrt{17}} \]
Thus, we can write:
\[ I = \frac{2}{\sqrt{17}} \cdot \frac{\sqrt{17}}{2} \sin^{-1}\left(\frac{2}{\sqrt{17}} \left(x + \frac{3}{2}\right)\right) + C \]
This simplifies to:
\[ I = \sin^{-1}\left(\frac{2}{\sqrt{17}} \left(x + \frac{3}{2}\right)\right) + C \]

### Step 5: Identify \( f(g(x)) \)
From the expression \( I = \sin^{-1}\left(\frac{2}{\sqrt{17}} \left(x + \frac{3}{2}\right)\right) + C \), we can identify:
- \( g(x) = \frac{2}{\sqrt{17}}(x + \frac{3}{2}) \)
- \( f(x) = \sin^{-1}(x) \)

### Final Answer
Thus, we have:
\[ I = f(g(x)) + C \]
where \( f(x) = \sin^{-1}(x) \) and \( g(x) = \frac{2}{\sqrt{17}}(x + \frac{3}{2}) \).