Evaluate the following integrals.
` int (2x -3)/(sqrt(2x-1))dx `

Answer

To evaluate the integral \( I = \int \frac{2x - 3}{\sqrt{2x - 1}} \, dx \), we will use the substitution method. Here are the steps:

### Step 1: Substitution
Let \( t^2 = 2x - 1 \). Then, differentiating both sides gives:
\[ 2 \, dx = 2t \, dt \quad \Rightarrow \quad dx = t \, dt \]

### Step 2: Express \( x \) in terms of \( t \)
From the substitution \( t^2 = 2x - 1 \), we can express \( x \) as:
\[ 2x = t^2 + 1 \quad \Rightarrow \quad x = \frac{t^2 + 1}{2} \]

### Step 3: Substitute in the integral
Now substitute \( x \) and \( dx \) into the integral:
\[ I = \int \frac{2\left(\frac{t^2 + 1}{2}\right) - 3}{\sqrt{t^2}} \cdot t \, dt \]
This simplifies to:
\[ I = \int \frac{t^2 + 1 - 6}{t} \cdot t \, dt = \int (t^2 - 5) \, dt \]

### Step 4: Integrate
Now we can integrate:
\[ I = \int t^2 \, dt - 5 \int dt = \frac{t^3}{3} - 5t + C \]

### Step 5: Substitute back for \( t \)
Recall that \( t = \sqrt{2x - 1} \). Substitute back:
\[ I = \frac{(\sqrt{2x - 1})^3}{3} - 5\sqrt{2x - 1} + C \]
This simplifies to:
\[ I = \frac{(2x - 1)^{3/2}}{3} - 5(2x - 1)^{1/2} + C \]

### Final Answer
Thus, the evaluated integral is:
\[ I = \frac{(2x - 1)^{3/2}}{3} - 5(2x - 1)^{1/2} + C \]