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

Answer

To solve the integral \( \int \frac{x}{\sqrt{x^2 + x + 1}} \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the integrand
We want to express \( x \) in terms of the derivative of \( x^2 + x + 1 \). Let's assume:
\[ x = a \cdot \frac{d}{dx}(x^2 + x + 1) + b \]
Calculating the derivative:
\[ \frac{d}{dx}(x^2 + x + 1) = 2x + 1 \]
Thus, we can write:
\[ x = a(2x + 1) + b \]

### Step 2: Compare coefficients
Now, we compare coefficients from both sides:
- Coefficient of \( x \): \( 1 = 2a \) → \( a = \frac{1}{2} \)
- Constant term: \( 0 = a + b \) → \( b = -\frac{1}{2} \)

### Step 3: Substitute back into the integral
Now substituting \( a \) and \( b \) back, we have:
\[ x = \frac{1}{2}(2x + 1) - \frac{1}{2} \]
Thus, we can rewrite the integral:
\[ \int \frac{x}{\sqrt{x^2 + x + 1}} \, dx = \int \frac{\frac{1}{2}(2x + 1) - \frac{1}{2}}{\sqrt{x^2 + x + 1}} \, dx \]
This simplifies to:
\[ \frac{1}{2} \int \frac{2x + 1}{\sqrt{x^2 + x + 1}} \, dx - \frac{1}{2} \int \frac{1}{\sqrt{x^2 + x + 1}} \, dx \]

### Step 4: Change of variable
Let \( t = x^2 + x + 1 \). Then, the derivative is:
\[ dt = (2x + 1) \, dx \implies dx = \frac{dt}{2x + 1} \]
Substituting this into the first integral:
\[ \frac{1}{2} \int \frac{dt}{\sqrt{t}} = \frac{1}{2} \cdot 2\sqrt{t} = \sqrt{t} = \sqrt{x^2 + x + 1} \]

### Step 5: Solve the second integral
For the second integral:
\[ -\frac{1}{2} \int \frac{1}{\sqrt{x^2 + x + 1}} \, dx \]
To solve this, we can complete the square:
\[ x^2 + x + 1 = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4} \]
Thus:
\[ -\frac{1}{2} \int \frac{1}{\sqrt{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}} \, dx \]
Using the formula for the integral of \( \frac{1}{\sqrt{x^2 + a^2}} \):
\[ = -\frac{1}{2} \log\left(x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right) + C \]

### Final Result
Combining both parts, we have:
\[ \int \frac{x}{\sqrt{x^2 + x + 1}} \, dx = \sqrt{x^2 + x + 1} - \frac{1}{2} \log\left(x + \frac{1}{2} + \sqrt{x^2 + x + 1}\right) + C \]