`int(1)/(2x^(2)+5x+3)dx`

Answer

To solve the integral \( \int \frac{1}{2x^2 + 5x + 3} \, dx \), we will follow these steps:

### Step 1: Rewrite the Integral
Let
\[ I = \int \frac{1}{2x^2 + 5x + 3} \, dx \]

### Step 2: Factor Out the Constant
We can factor out \( \frac{1}{2} \) from the denominator:
\[ I = \frac{1}{2} \int \frac{1}{x^2 + \frac{5}{2}x + \frac{3}{2}} \, dx \]

### Step 3: Complete the Square
Next, we need to complete the square for the quadratic expression in the denominator:
\[ x^2 + \frac{5}{2}x + \frac{3}{2} \]
To complete the square, we take half of the coefficient of \( x \) (which is \( \frac{5}{4} \)), square it (which gives \( \frac{25}{16} \)), and adjust the expression:
\[ x^2 + \frac{5}{2}x + \frac{3}{2} = \left(x + \frac{5}{4}\right)^2 - \frac{25}{16} + \frac{24}{16} = \left(x + \frac{5}{4}\right)^2 - \frac{1}{16} \]

### Step 4: Substitute Back into the Integral
Now we substitute this back into the integral:
\[ I = \frac{1}{2} \int \frac{1}{\left(x + \frac{5}{4}\right)^2 - \left(\frac{1}{4}\right)^2} \, dx \]

### Step 5: Use the Formula for the Integral of a Difference of Squares
The integral can be solved using the formula:
\[ \int \frac{1}{u^2 - a^2} \, du = \frac{1}{2a} \ln \left| \frac{u - a}{u + a} \right| + C \]
where \( u = x + \frac{5}{4} \) and \( a = \frac{1}{4} \).

### Step 6: Apply the Formula
Applying the formula:
\[ I = \frac{1}{2} \cdot \frac{1}{2 \cdot \frac{1}{4}} \ln \left| \frac{x + \frac{5}{4} - \frac{1}{4}}{x + \frac{5}{4} + \frac{1}{4}} \right| + C \]
This simplifies to:
\[ I = \frac{1}{2} \cdot 2 \ln \left| \frac{x + 1}{x + \frac{3}{2}} \right| + C \]
Thus,
\[ I = \ln \left| \frac{x + 1}{x + \frac{3}{2}} \right| + C \]

### Final Answer
The final answer is:
\[ \int \frac{1}{2x^2 + 5x + 3} \, dx = \ln \left| \frac{2x + 2}{2x + 3} \right| + C \]