The value of `int(1)/((2x-1)sqrt(x^(2)-x))dx` is equal to (where c is the constant of integration)

Options

A. `sec^(-1)(x-1)+c`
B. `sec^(-1)(2x-1)+c`
C. `tan^(-1)x+c`
D. `tan^(-1)(2x-1)+c`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{1}{(2x-1) \sqrt{x^2 - x}} \, dx \]

### Step 2: Simplify the Square Root
Notice that \( x^2 - x = x(x-1) \). Thus, we can rewrite the integral as:
\[ I = \int \frac{1}{(2x-1) \sqrt{x(x-1)}} \, dx \]

### Step 3: Use a Substitution
Let \( u = 2x - 1 \). Then, \( du = 2 \, dx \) or \( dx = \frac{du}{2} \). Also, we can express \( x \) in terms of \( u \):
\[ x = \frac{u + 1}{2} \]
Substituting \( x \) into \( \sqrt{x(x-1)} \):
\[ x - 1 = \frac{u + 1}{2} - 1 = \frac{u - 1}{2} \]
Thus,
\[ \sqrt{x(x-1)} = \sqrt{\frac{u + 1}{2} \cdot \frac{u - 1}{2}} = \frac{1}{2} \sqrt{(u + 1)(u - 1)} = \frac{1}{2} \sqrt{u^2 - 1} \]

### Step 4: Substitute into the Integral
Now substitute \( u \) and \( dx \) into the integral:
\[ I = \int \frac{1}{u \cdot \frac{1}{2} \sqrt{u^2 - 1}} \cdot \frac{du}{2} \]
This simplifies to:
\[ I = \int \frac{1}{u \sqrt{u^2 - 1}} \, du \]

### Step 5: Recognize the Integral
The integral \( \int \frac{1}{u \sqrt{u^2 - 1}} \, du \) is known to be:
\[ \int \frac{1}{u \sqrt{u^2 - 1}} \, du = \sec^{-1}(u) + C \]

### Step 6: Substitute Back
Now substitute back \( u = 2x - 1 \):
\[ I = \sec^{-1}(2x - 1) + C \]

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

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