`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 given by the function:

\[ f(x) = \int \frac{x^2 + x + 1}{x + 1 + \sqrt{x}} \, dx \]

We want to find \( f(1) \).

### Step-by-Step Solution:

1. **Rewrite the Numerator:**
We can express the numerator \( x^2 + x + 1 \) in a different way. Notice that we can factor it as follows:
\[ x^2 + x + 1 = (x + 1 + \sqrt{x})(x + 1 - \sqrt{x}) \]
This is because:
\[ (x + 1 + \sqrt{x})(x + 1 - \sqrt{x}) = (x + 1)^2 - (\sqrt{x})^2 = x^2 + 2x + 1 - x = x^2 + x + 1 \]

2. **Substitute into the Integral:**
Now we can substitute this back into the integral:
\[ f(x) = \int \frac{(x + 1 + \sqrt{x})(x + 1 - \sqrt{x})}{x + 1 + \sqrt{x}} \, dx \]
The \( x + 1 + \sqrt{x} \) terms cancel out:
\[ f(x) = \int (x + 1 - \sqrt{x}) \, dx \]

3. **Integrate Each Term:**
Now we can integrate term by term:
\[ f(x) = \int (x + 1 - \sqrt{x}) \, dx = \int x \, dx + \int 1 \, dx - \int \sqrt{x} \, dx \]
\[ = \frac{x^2}{2} + x - \frac{2}{3} x^{3/2} + C \]

4. **Evaluate \( f(1) \):**
Now we substitute \( x = 1 \) into the integrated function:
\[ f(1) = \frac{1^2}{2} + 1 - \frac{2}{3} (1)^{3/2} + C \]
\[ = \frac{1}{2} + 1 - \frac{2}{3} + C \]
\[ = \frac{1}{2} + \frac{2}{2} - \frac{2}{3} + C \]
\[ = \frac{3}{2} - \frac{2}{3} + C \]

5. **Finding a Common Denominator:**
To combine these fractions, we find a common denominator (which is 6):
\[ = \frac{9}{6} - \frac{4}{6} + C = \frac{5}{6} + C \]

6. **Final Result:**
Since we do not have a specific value for \( C \) (the constant of integration), we can conclude:
\[ f(1) = \frac{5}{6} + C \]
If \( C \) is assumed to be 0 (as is often the case when evaluating definite integrals), then:
\[ f(1) = \frac{5}{6} \]

### Conclusion:
Thus, the final answer for \( f(1) \) is:
\[ \boxed{\frac{5}{6}} \]