The value of `int(dx)/(x^(2)+x+1)` is equal to

Options

A. `sqrt(3)/(2)tan^(-1)(2x+1)/(sqrt(3))C`
B. `2/sqrt(3)tan^(-1)(2x+1)/(sqrt(3))+C`
C. `1/sqrt(3)tan^(-1)(2x+1)/(sqrt(3))+C`
D. `2/sqrt(3)tan^(-1)(2x-1)/(sqrt(3))+C`

Answer

To solve the integral \(\int \frac{dx}{x^2 + x + 1}\), we will use the method of completing the square and then apply the formula for the integral of a function of the form \(\frac{1}{x^2 + a^2}\).

### Step 1: Complete the square for the denominator
We start with the quadratic expression in the denominator:
\[ x^2 + x + 1 \]
To complete the square, we take the coefficient of \(x\), which is \(1\), halve it to get \(\frac{1}{2}\), and then square it to get \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\). We can rewrite the expression as:
\[ x^2 + x + \frac{1}{4} + \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]
Thus, we have:
\[ x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]

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

### Step 3: Use the formula for integration
The integral \(\int \frac{dx}{x^2 + a^2}\) is given by \(\frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C\). In our case, we have:
- \(a^2 = \frac{3}{4}\) which gives \(a = \frac{\sqrt{3}}{2}\).

Thus, we can apply the formula:
\[ \int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C \]

### Step 4: Simplify the expression
This simplifies to:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2\left(x + \frac{1}{2}\right)}{\sqrt{3}}\right) + C \]
Now substituting \(x + \frac{1}{2} = x + 0.5\) gives:
\[ \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right) + C \]

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