`int_(-1)^(1)(x^(2)+x)/(x^(2)+1)dx=`

Options

A. `2-(pi)/(2)`
B. `(pi)/(2)+2`
C. `(pi)/(2)-2`
D. 2

Answer

To solve the definite integral

\[ \int_{-1}^{1} \frac{x^2 + x}{x^2 + 1} \, dx, \]

we can break it down into simpler parts.

### Step 1: Split the Integral
We can separate the integral into two parts:

\[ \int_{-1}^{1} \frac{x^2 + x}{x^2 + 1} \, dx = \int_{-1}^{1} \frac{x^2}{x^2 + 1} \, dx + \int_{-1}^{1} \frac{x}{x^2 + 1} \, dx. \]

### Step 2: Evaluate the First Integral
Now, we evaluate the first integral:

\[ \int_{-1}^{1} \frac{x^2}{x^2 + 1} \, dx. \]

This can be simplified as:

\[ \int_{-1}^{1} \left( 1 - \frac{1}{x^2 + 1} \right) \, dx = \int_{-1}^{1} 1 \, dx - \int_{-1}^{1} \frac{1}{x^2 + 1} \, dx. \]

Calculating the first part:

\[ \int_{-1}^{1} 1 \, dx = [x]_{-1}^{1} = 1 - (-1) = 2. \]

### Step 3: Evaluate the Second Integral
Now, we need to evaluate the second integral:

\[ \int_{-1}^{1} \frac{1}{x^2 + 1} \, dx. \]

This integral is known and can be computed as:

\[ \int \frac{1}{x^2 + 1} \, dx = \tan^{-1}(x). \]

Thus,

\[ \int_{-1}^{1} \frac{1}{x^2 + 1} \, dx = \left[ \tan^{-1}(x) \right]_{-1}^{1} = \tan^{-1}(1) - \tan^{-1}(-1) = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}. \]

### Step 4: Combine the Results
Now, we can combine the results from Step 2 and Step 3:

\[ \int_{-1}^{1} \frac{x^2}{x^2 + 1} \, dx = 2 - \frac{\pi}{2}. \]

### Step 5: Evaluate the Second Integral (Odd Function)
Next, we evaluate the second integral:

\[ \int_{-1}^{1} \frac{x}{x^2 + 1} \, dx. \]

Since \(\frac{x}{x^2 + 1}\) is an odd function, the integral over a symmetric interval around zero is zero:

\[ \int_{-1}^{1} \frac{x}{x^2 + 1} \, dx = 0. \]

### Final Result
Combining everything, we have:

\[ \int_{-1}^{1} \frac{x^2 + x}{x^2 + 1} \, dx = \left(2 - \frac{\pi}{2}\right) + 0 = 2 - \frac{\pi}{2}. \]

Thus, the final answer is:

\[ \int_{-1}^{1} \frac{x^2 + x}{x^2 + 1} \, dx = 2 - \frac{\pi}{2}. \]

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