`int(dx)/((2x^(2)+x+3))=?`

Options

A. `(1)/(sqrt(23))tan^(-1)((4x+1)/(sqrt(23)))+C`
B. `(1)/(sqrt(23))tan^(-1)((x+1)/(sqrt(23)))+C`
C. `(2)/(sqrt(23))tan^(-1)((4x+1)/(sqrt(23)))+C`
D. None of these

Explanation

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

### Step 1: Factor out the constant
First, we can factor out \( \frac{1}{2} \) from the integral:
\[ \int \frac{dx}{2x^2 + x + 3} = \frac{1}{2} \int \frac{dx}{x^2 + \frac{1}{2}x + \frac{3}{2}} \]

### Step 2: Complete the square
Next, we need to complete the square for the quadratic expression in the denominator:
\[ x^2 + \frac{1}{2}x + \frac{3}{2} \]
To complete the square, we take half of the coefficient of \(x\) (which is \(\frac{1}{2}\)), square it (which gives \(\frac{1}{16}\)), and rewrite the expression:
\[ x^2 + \frac{1}{2}x + \frac{3}{2} = \left(x + \frac{1}{4}\right)^2 - \frac{1}{16} + \frac{3}{2} \]
Now, convert \(\frac{3}{2}\) to have a common denominator:
\[ \frac{3}{2} = \frac{24}{16} \]
Thus, we have:
\[ \left(x + \frac{1}{4}\right)^2 + \left(\frac{24}{16} - \frac{1}{16}\right) = \left(x + \frac{1}{4}\right)^2 + \frac{23}{16} \]

### Step 3: Rewrite the integral
Now we can rewrite the integral:
\[ \frac{1}{2} \int \frac{dx}{\left(x + \frac{1}{4}\right)^2 + \frac{23}{16}} \]

### Step 4: Use the formula for integration
We can use the formula for the integral of the form \(\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\). Here, \(a^2 = \frac{23}{16}\), so \(a = \frac{\sqrt{23}}{4}\).

Thus, we have:
\[ \frac{1}{2} \cdot \frac{4}{\sqrt{23}} \tan^{-1}\left(\frac{x + \frac{1}{4}}{\frac{\sqrt{23}}{4}}\right) + C \]

### Step 5: Simplify the expression
This simplifies to:
\[ \frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{4(x + \frac{1}{4})}{\sqrt{23}}\right) + C \]
\[ = \frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \]

### Final Answer
Thus, the final result of the integral is:
\[ \int \frac{dx}{2x^2 + x + 3} = \frac{2}{\sqrt{23}} \tan^{-1}\left(\frac{4x + 1}{\sqrt{23}}\right) + C \]