`int(1)/(sqrt(2x^(2)+3x-2))dx`

Answer

To solve the integral \( \int \frac{1}{\sqrt{2x^2 + 3x - 2}} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
Let \( I = \int \frac{1}{\sqrt{2x^2 + 3x - 2}} \, dx \).

### Step 2: Factor Out the Coefficient of \( x^2 \)
We can factor out 2 from the quadratic expression in the denominator:
\[ I = \int \frac{1}{\sqrt{2 \left( x^2 + \frac{3}{2}x - 1 \right)}} \, dx = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{x^2 + \frac{3}{2}x - 1}} \, dx \]

### Step 3: Complete the Square
Next, we need to complete the square for the expression \( x^2 + \frac{3}{2}x - 1 \):
\[ x^2 + \frac{3}{2}x - 1 = \left( x + \frac{3}{4} \right)^2 - \left( \frac{9}{16} + 1 \right) = \left( x + \frac{3}{4} \right)^2 - \frac{25}{16} \]
Thus, we rewrite the integral:
\[ I = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{\left( x + \frac{3}{4} \right)^2 - \left( \frac{5}{4} \right)^2}} \, dx \]

### Step 4: Use the Integral Formula
We can use the formula:
\[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln |x + \sqrt{x^2 - a^2}| + C \]
In our case, \( x \) is \( x + \frac{3}{4} \) and \( a \) is \( \frac{5}{4} \):
\[ I = \frac{1}{\sqrt{2}} \left( \ln \left| x + \frac{3}{4} + \sqrt{\left( x + \frac{3}{4} \right)^2 - \left( \frac{5}{4} \right)^2} \right| + C \right) \]

### Step 5: Simplify the Expression
Now we simplify the expression:
\[ I = \frac{1}{\sqrt{2}} \left( \ln \left| x + \frac{3}{4} + \sqrt{(x + \frac{3}{4})^2 - \frac{25}{16}} \right| + C \right) \]

### Final Result
Thus, the final result of the integral is:
\[ I = \frac{1}{\sqrt{2}} \ln \left| x + \frac{3}{4} + \sqrt{(x + \frac{3}{4})^2 - \frac{25}{16}} \right| + C \]