The value of `int(dx)/(sqrt(4x-3-x^(2)))` is equal to

Options

A. `sin^(-1) (x - 1) + C`
B. `log|(x-2)+sqrt(4x-3-x^(2))|`
C. `log|(x-1)+sqrt(4x-3-x^(2))|+C`
D. `sin^(-1)(x-2)+C`

Answer

To solve the integral \(\int \frac{dx}{\sqrt{4x - 3 - x^2}}\), we will follow these steps:

### Step 1: Rewrite the integrand
First, we rewrite the expression under the square root to make it easier to integrate. We have:
\[ 4x - 3 - x^2 = - (x^2 - 4x + 3) \]
Next, we complete the square for the quadratic \(x^2 - 4x + 3\):
\[ x^2 - 4x + 3 = (x - 2)^2 - 1 \]
Thus, we can rewrite the integral as:
\[ \int \frac{dx}{\sqrt{-( (x - 2)^2 - 1 )}} = \int \frac{dx}{\sqrt{1 - (x - 2)^2}} \]

### Step 2: Substitute to simplify the integral
Let \(t = x - 2\). Then, \(dx = dt\), and the integral becomes:
\[ \int \frac{dt}{\sqrt{1 - t^2}} \]

### Step 3: Recognize the integral
The integral \(\int \frac{dt}{\sqrt{1 - t^2}}\) is a standard integral, which is equal to:
\[ \sin^{-1}(t) + C \]

### Step 4: Substitute back to original variable
Now, we substitute back \(t = x - 2\):
\[ \sin^{-1}(x - 2) + C \]

### Final Answer
Thus, the value of the integral \(\int \frac{dx}{\sqrt{4x - 3 - x^2}}\) is:
\[ \sin^{-1}(x - 2) + C \]

---