1. Key Concepts and Formulas

• Definite Integrals of Odd and Even Functions: For a definite integral over a symmetric interval $[-a, a]$:
  • If $f(x)$ is an odd function (i.e., $f(-x) = -f(x)$ for all $x$ in the domain), then $\int_{-a}^a f(x) dx = 0$
  • If $f(x)$ is an even function (i.e., $f(-x) = f(x)$ for all $x$ in the domain), then $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$
• Integration by Parts: The formula for integration by parts is $\int u \, dv = uv - \int v \, du$
• Logarithmic Properties: The natural logarithm of 1 is 0, i.e., $\log(1) = 0$.

2. Step-by-Step Solution

Step 1: Analyze the integrand and the interval of integration. The integral is given by $\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx}$. The interval of integration is $[-1, 1]$, which is a symmetric interval about 0. This suggests we should check if the integrand is an odd or even function.

Step 2: Determine if the integrand is an odd or even function. Let the integrand be $f(x) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$. We need to evaluate $f(-x)$. $f(-x) = \log \left( {-x + \sqrt {{(-x)^2} + 1} } \right)$ $f(-x) = \log \left( {-x + \sqrt {{x^2} + 1} } \right)$ To simplify this expression, we can multiply the argument of the logarithm by $\frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1} + x}$ (which is equal to 1). $f(-x) = \log \left( \frac{{\left( {\sqrt {{x^2} + 1} - x} \right)\left( {\sqrt {{x^2} + 1} + x} \right)}}{{\sqrt {{x^2} + 1} + x}} \right)$ Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, where $a = \sqrt{x^2+1}$ and $b=x$: $f(-x) = \log \left( \frac{{(\sqrt {{x^2} + 1})^2 - x^2}}{{\sqrt {{x^2} + 1} + x}} \right)$ $f(-x) = \log \left( \frac{{(x^2 + 1) - x^2}}{{\sqrt {{x^2} + 1} + x}} \right)$ $f(-x) = \log \left( \frac{1}{{\sqrt {{x^2} + 1} + x}} \right)$ Using the logarithm property $\log \left( \frac{1}{a} \right) = -\log(a)$: $f(-x) = -\log \left( {\sqrt {{x^2} + 1} + x} \right)$ So, we have $f(-x) = -f(x)$. This means that the integrand $f(x)$ is an odd function.

Step 3: Apply the property of definite integrals for odd functions. Since the integrand $f(x) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$ is an odd function and the interval of integration is symmetric $[-1, 1]$, we can directly apply the property: $\int_{-a}^a f(x) dx = 0$ In this case, $a=1$, so: $\int\limits_{ - 1}^1 {\log \left( {x + \sqrt {{x^2} + 1} } \right)dx} = 0$

3. Common Mistakes & Tips

• Algebraic Errors: Be very careful with algebraic manipulations, especially when dealing with square roots and logarithms. A small error can lead to an incorrect conclusion about the function's parity.
• Forgetting the Symmetric Interval: The property $\int_{-a}^a f(x) dx = 0$ for odd functions only applies when the interval is symmetric around zero. If the interval were, for example, $[-1, 2]$, this property could not be used directly.
• Direct Integration: While it's possible to solve this integral using integration by parts, it's significantly more complex and time-consuming than recognizing the odd nature of the integrand. Always look for symmetry properties first for integrals over symmetric intervals.

4. Summary

The problem requires evaluating a definite integral over a symmetric interval $[-1, 1]$. By analyzing the integrand, $f(x) = \log \left( {x + \sqrt {{x^2} + 1} } \right)$, we determined that it is an odd function, meaning $f(-x) = -f(x)$. For definite integrals of odd functions over symmetric intervals of the form $[-a, a]$, the value of the integral is always 0. Therefore, the given integral evaluates to 0.

5. Final Answer

The final answer is \boxed{0}.