Evaluate `int_(-1)^(1)(x^7+tan^5x + x+ 1) dx`

Answer

To evaluate the definite integral

\[ \int_{-1}^{1} (x^7 + \tan^5 x + x + 1) \, dx, \]

we will break it down into manageable parts and utilize properties of odd and even functions.

### Step 1: Break down the integral

We can separate the integral into four parts:

\[ \int_{-1}^{1} (x^7 + \tan^5 x + x + 1) \, dx = \int_{-1}^{1} x^7 \, dx + \int_{-1}^{1} \tan^5 x \, dx + \int_{-1}^{1} x \, dx + \int_{-1}^{1} 1 \, dx. \]

### Step 2: Evaluate each integral

1. **Evaluate \(\int_{-1}^{1} x^7 \, dx\)**:
- The function \(x^7\) is an odd function because \(f(-x) = -f(x)\). Therefore, the integral over a symmetric interval around zero is zero:
\[ \int_{-1}^{1} x^7 \, dx = 0. \]

2. **Evaluate \(\int_{-1}^{1} \tan^5 x \, dx\)**:
- The function \(\tan^5 x\) is also an odd function. Thus, this integral is also zero:
\[ \int_{-1}^{1} \tan^5 x \, dx = 0. \]

3. **Evaluate \(\int_{-1}^{1} x \, dx\)**:
- The function \(x\) is an odd function, so this integral is also zero:
\[ \int_{-1}^{1} x \, dx = 0. \]

4. **Evaluate \(\int_{-1}^{1} 1 \, dx\)**:
- The integral of 1 over the interval from -1 to 1 is simply the length of the interval:
\[ \int_{-1}^{1} 1 \, dx = 1 - (-1) = 2. \]

### Step 3: Combine the results

Now we can combine the results of all four integrals:

\[ \int_{-1}^{1} (x^7 + \tan^5 x + x + 1) \, dx = 0 + 0 + 0 + 2 = 2. \]

### Final Answer

Thus, the value of the integral is

\[ \boxed{2}. \]