`int(dx)/(sqrt(3+4x-4x^2))=`

Options

A. `(1)/(2)cos^(-1)((2x-1)/(2))+c`
B. `2cos^(-1)((2x-1)/(2))+c`
C. `(1)/(2)sin^(-1)((2x-1)/(2))+c`
D. `2sin^(-1)((2x-1)/(2))+c`

Answer

To solve the integral

\[ I = \int \frac{dx}{\sqrt{3 + 4x - 4x^2}}, \]

we will follow these steps:

### Step 1: Rewrite the expression under the square root
First, we rewrite the expression inside the square root:

\[ 3 + 4x - 4x^2 = -4(x^2 - x - \frac{3}{4}). \]

### Step 2: Factor out the negative sign
Next, we factor out the negative sign from the quadratic expression:

\[ I = \int \frac{dx}{\sqrt{-4(x^2 - x - \frac{3}{4})}} = \int \frac{dx}{2\sqrt{-(x^2 - x - \frac{3}{4})}}. \]

### Step 3: Complete the square
Now, we complete the square for the quadratic expression \(x^2 - x - \frac{3}{4}\):

\[ x^2 - x - \frac{3}{4} = \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} - \frac{3}{4} = \left(x - \frac{1}{2}\right)^2 - 1. \]

### Step 4: Substitute back into the integral
Substituting this back into the integral gives us:

\[ I = \int \frac{dx}{2\sqrt{-4\left(\left(x - \frac{1}{2}\right)^2 - 1\right)}} = \int \frac{dx}{2\sqrt{-4\left(\left(x - \frac{1}{2}\right)^2 - 1\right)}}. \]

### Step 5: Simplify the integral
This can be simplified to:

\[ I = \int \frac{dx}{2\sqrt{4 - 4\left(x - \frac{1}{2}\right)^2}} = \frac{1}{2} \int \frac{dx}{\sqrt{1 - \left(x - \frac{1}{2}\right)^2}}. \]

### Step 6: Use the standard integral formula
The integral \(\int \frac{dx}{\sqrt{1 - u^2}} = \sin^{-1}(u) + C\) can be applied here, where \(u = x - \frac{1}{2}\):

\[ I = \frac{1}{2} \sin^{-1}\left(x - \frac{1}{2}\right) + C. \]

### Step 7: Final answer
Thus, the final answer is:

\[ I = \frac{1}{2} \sin^{-1}(2x - 1) + C. \]

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