`int(1)/(sqrt(8+2x-x^(2)))dx` is equal to

Options

A. `(1)/(3)sin^(-1)((x-1)/(3))+c`
B. `sin^(-1)((x+1)/(3))+c`
C. `(1)/(3)sin^(-1)((x+1)/(3))+c`
D. `sin^(-1)((x-1)/(3))+c`

Explanation

To solve the integral \( \int \frac{1}{\sqrt{8 + 2x - x^2}} \, dx \), we can follow these steps:

### Step 1: Rewrite the expression under the square root
We start by rewriting the expression inside the square root:
\[ 8 + 2x - x^2 = -(x^2 - 2x - 8) \]
Now, we can complete the square for the quadratic expression \( - (x^2 - 2x - 8) \).

### Step 2: Completing the square
To complete the square for \( x^2 - 2x - 8 \):
\[ x^2 - 2x = (x - 1)^2 - 1 \]
Thus,
\[ x^2 - 2x - 8 = (x - 1)^2 - 9 \]
So,
\[ 8 + 2x - x^2 = -((x - 1)^2 - 9) = 9 - (x - 1)^2 \]

### Step 3: Substitute back into the integral
Now we can rewrite the integral:
\[ \int \frac{1}{\sqrt{9 - (x - 1)^2}} \, dx \]

### Step 4: Use the standard integral formula
We recognize that the integral \( \int \frac{1}{\sqrt{a^2 - u^2}} \, du = \sin^{-1} \left( \frac{u}{a} \right) + C \). Here, \( a = 3 \) and \( u = x - 1 \).

### Step 5: Apply the formula
Thus, we can apply the formula:
\[ \int \frac{1}{\sqrt{9 - (x - 1)^2}} \, dx = \sin^{-1} \left( \frac{x - 1}{3} \right) + C \]

### Final Answer
The final answer for the integral is:
\[ \int \frac{1}{\sqrt{8 + 2x - x^2}} \, dx = \sin^{-1} \left( \frac{x - 1}{3} \right) + C \]