`int_(0)^(2) (1)/(4+x+x^(2))dx`

Answer

To evaluate the integral \( I = \int_{0}^{2} \frac{1}{4 + x + x^2} \, dx \), we will follow these steps:

### Step 1: Rewrite the Denominator
We start by rewriting the denominator \( 4 + x + x^2 \). We can complete the square for the quadratic expression \( x^2 + x + 4 \).

\[ x^2 + x + 4 = \left( x + \frac{1}{2} \right)^2 + \left( 4 - \frac{1}{4} \right) = \left( x + \frac{1}{2} \right)^2 + \frac{15}{4} \]

### Step 2: Substitute in the Integral
Now we substitute this back into the integral:

\[ I = \int_{0}^{2} \frac{1}{\left( x + \frac{1}{2} \right)^2 + \frac{15}{4}} \, dx \]

### Step 3: Use a Trigonometric Substitution
To evaluate this integral, we can use the substitution \( u = x + \frac{1}{2} \), which gives \( du = dx \). The limits change as follows:
- When \( x = 0 \), \( u = \frac{1}{2} \)
- When \( x = 2 \), \( u = 2 + \frac{1}{2} = \frac{5}{2} \)

Thus, we have:

\[ I = \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{1}{u^2 + \frac{15}{4}} \, du \]

### Step 4: Factor Out the Constant
We can factor out \( \frac{15}{4} \) from the denominator:

\[ I = \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{1}{\frac{15}{4} \left( \frac{4u^2}{15} + 1 \right)} \, du = \frac{4}{15} \int_{\frac{1}{2}}^{\frac{5}{2}} \frac{1}{\frac{4u^2}{15} + 1} \, du \]

### Step 5: Use the Arctangent Formula
The integral of \( \frac{1}{a^2 + u^2} \) is \( \frac{1}{a} \tan^{-1} \left( \frac{u}{a} \right) \). Here, \( a^2 = 1 \) and \( a = \sqrt{\frac{15}{4}} = \frac{\sqrt{15}}{2} \).

Thus, we have:

\[ I = \frac{4}{15} \cdot \frac{2}{\sqrt{15}} \left[ \tan^{-1} \left( \frac{u}{\frac{\sqrt{15}}{2}} \right) \right]_{\frac{1}{2}}^{\frac{5}{2}} \]

### Step 6: Evaluate the Limits
Now we evaluate the limits:

\[ I = \frac{8}{15\sqrt{15}} \left[ \tan^{-1} \left( \frac{5}{\sqrt{15}} \right) - \tan^{-1} \left( \frac{1}{\sqrt{15}} \right) \right] \]

### Step 7: Final Calculation
We can simplify the arctangent terms, but the exact values depend on the calculator or tables. Thus, we can leave it in this form or compute the numerical values.

### Final Result
The final value of the integral is:

\[ I = \frac{8}{15\sqrt{15}} \left[ \tan^{-1} \left( \frac{5}{\sqrt{15}} \right) - \tan^{-1} \left( \frac{1}{\sqrt{15}} \right) \right] \]