`f(x) = int(x^(2)+x+1)/(x+1+sqrt(x))dx`, then f(1) =

Options

A. `(7)/(6)`
B. `(2)/(3)`
C. `(11)/(6)`
D. `(5)/(6)`

Explanation

To solve the problem, we need to evaluate the integral \( f(x) = \int \frac{x^2 + x + 1}{x + 1 + \sqrt{x}} \, dx \) and then find \( f(1) \).

### Step-by-Step Solution:

1. **Rewrite the Integral**:
We start with the integral:
\[ f(x) = \int \frac{x^2 + x + 1}{x + 1 + \sqrt{x}} \, dx \]

2. **Factor the Numerator**:
We can use the expansion formula mentioned in the video:
\[ x^2 + x + 1 = (x + 1 + \sqrt{x})(x + 1 - \sqrt{x}) \]
Thus, we can rewrite the integral as:
\[ f(x) = \int \frac{(x + 1 + \sqrt{x})(x + 1 - \sqrt{x})}{x + 1 + \sqrt{x}} \, dx \]

3. **Cancel the Denominator**:
The \( x + 1 + \sqrt{x} \) in the numerator and denominator cancels out:
\[ f(x) = \int (x + 1 - \sqrt{x}) \, dx \]

4. **Integrate Each Term**:
Now we can integrate each term separately:
\[ f(x) = \int x \, dx + \int 1 \, dx - \int \sqrt{x} \, dx \]
- The integral of \( x \) is \( \frac{x^2}{2} \).
- The integral of \( 1 \) is \( x \).
- The integral of \( \sqrt{x} \) is \( \frac{2}{3} x^{3/2} \).

Thus,
\[ f(x) = \frac{x^2}{2} + x - \frac{2}{3} x^{3/2} + C \]

5. **Evaluate \( f(1) \)**:
Now we substitute \( x = 1 \):
\[ f(1) = \frac{1^2}{2} + 1 - \frac{2}{3} \cdot 1^{3/2} \]
Simplifying this gives:
\[ f(1) = \frac{1}{2} + 1 - \frac{2}{3} \]
Converting to a common denominator (6):
\[ f(1) = \frac{3}{6} + \frac{6}{6} - \frac{4}{6} = \frac{5}{6} \]

6. **Final Answer**:
Therefore, the value of \( f(1) \) is:
\[ f(1) = \frac{5}{6} \]