`int(2x-1)/(sqrt(x^(2)-x-1))dx`

Answer

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

### Step 1: Set up the integral
Let:
\[ I = \int \frac{2x - 1}{\sqrt{x^2 - x - 1}} \, dx \]

### Step 2: Substitute for the denominator
We will let \( t = x^2 - x - 1 \). Now, we need to differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 2x - 1 \]
Thus, we can express \( dx \) in terms of \( dt \):
\[ dt = (2x - 1) \, dx \implies dx = \frac{dt}{2x - 1} \]

### Step 3: Substitute in the integral
Substituting \( t \) and \( dx \) into the integral gives:
\[ I = \int \frac{dt}{\sqrt{t}} \]

### Step 4: Simplify the integral
The integral can be rewritten as:
\[ I = \int t^{-\frac{1}{2}} \, dt \]

### Step 5: Integrate
Now we integrate:
\[ I = \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + C = 2t^{\frac{1}{2}} + C \]

### Step 6: Substitute back for \( t \)
Now we substitute back \( t = x^2 - x - 1 \):
\[ I = 2\sqrt{x^2 - x - 1} + C \]

### Final Answer
Thus, the solution to the integral is:
\[ \int \frac{2x - 1}{\sqrt{x^2 - x - 1}} \, dx = 2\sqrt{x^2 - x - 1} + C \]

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