Find the value of `int_(-1)^1[x^2+{x}]dx ,w h e r e[dot]a n d{dot}` denote the greatest function and fractional parts of `x ,` respectively.

Explanation

To solve the integral \( I = \int_{-1}^{1} \left[ x^2 + \{ x \} \right] dx \), where \([ \cdot ]\) denotes the greatest integer function and \(\{ \cdot \}\) denotes the fractional part of \(x\), we will break the integral into two parts based on the intervals of \(x\).

### Step 1: Break the Integral into Two Parts
The integral can be split into two intervals: from \(-1\) to \(0\) and from \(0\) to \(1\).

\[ I = \int_{-1}^{0} \left[ x^2 + \{ x \} \right] dx + \int_{0}^{1} \left[ x^2 + \{ x \} \right] dx \]

### Step 2: Evaluate the Integral from \(-1\) to \(0\)
In the interval \([-1, 0)\), the greatest integer function \([x]\) is \(-1\) and the fractional part \(\{x\} = x + 1\). Thus, we have:

\[ I_1 = \int_{-1}^{0} \left[ x^2 + (x + 1) \right] dx = \int_{-1}^{0} \left[ x^2 + x + 1 \right] dx \]

### Step 3: Simplify the Expression
Now, we can simplify the expression inside the integral:

\[ I_1 = \int_{-1}^{0} (x^2 + x + 1) dx \]

### Step 4: Calculate the Integral from \(-1\) to \(0\)
Now we calculate this integral:

\[ I_1 = \int_{-1}^{0} x^2 dx + \int_{-1}^{0} x dx + \int_{-1}^{0} 1 dx \]

Calculating each part:

1. \(\int_{-1}^{0} x^2 dx = \left[ \frac{x^3}{3} \right]_{-1}^{0} = 0 - \left( -\frac{1}{3} \right) = \frac{1}{3}\)
2. \(\int_{-1}^{0} x dx = \left[ \frac{x^2}{2} \right]_{-1}^{0} = 0 - \left( -\frac{1}{2} \right) = \frac{1}{2}\)
3. \(\int_{-1}^{0} 1 dx = [x]_{-1}^{0} = 0 - (-1) = 1\)

Adding these results together:

\[ I_1 = \frac{1}{3} + \frac{1}{2} + 1 = \frac{1}{3} + \frac{3}{6} + \frac{6}{6} = \frac{1}{3} + \frac{9}{6} = \frac{1}{3} + \frac{3}{2} = \frac{1}{3} + \frac{9}{6} = \frac{1 + 9}{6} = \frac{10}{6} = \frac{5}{3} \]

### Step 5: Evaluate the Integral from \(0\) to \(1\)
In the interval \([0, 1)\), the greatest integer function \([x]\) is \(0\) and the fractional part \(\{x\} = x\). Thus, we have:

\[ I_2 = \int_{0}^{1} \left[ x^2 + x \right] dx = \int_{0}^{1} (x^2 + x) dx \]

### Step 6: Calculate the Integral from \(0\) to \(1\)
Calculating this integral:

\[ I_2 = \int_{0}^{1} x^2 dx + \int_{0}^{1} x dx \]

Calculating each part:

1. \(\int_{0}^{1} x^2 dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1}{3} - 0 = \frac{1}{3}\)
2. \(\int_{0}^{1} x dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2}\)

Adding these results together:

\[ I_2 = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6} \]

### Step 7: Combine Both Parts
Now, we combine both parts:

\[ I = I_1 + I_2 = \frac{5}{3} + \frac{5}{6} \]

Finding a common denominator (which is 6):

\[ I = \frac{10}{6} + \frac{5}{6} = \frac{15}{6} = \frac{5}{2} \]

### Final Answer
Thus, the value of the integral is:

\[ \boxed{\frac{5}{2}} \]