Key Concepts and Formulas

• Definite Integral Property for Symmetric Intervals: For a function $f(x)$, $\int_{-a}^a f(x) \, dx = \int_0^a [f(x) + f(-x)] \, dx$. This property is particularly useful for integrals over symmetric intervals like $[-a, a]$ as it often simplifies the integrand.
• Definition of Absolute Value: $|x| = x$ if $x \ge 0$, and $|x| = -x$ if $x < 0$. This definition is crucial when dealing with absolute values within an integral, especially when the integration interval changes.
• Power Rule for Integration: $\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$.

Step-by-Step Solution

Step 1: Define the Integrand and Apply the Symmetry Property Let the given integral be $I$. $I = \int_{-1}^1 \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} \, dx$ Let $f(x) = \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}}$. We will use the property $\int_{-a}^a f(x) \, dx = \int_0^a [f(x) + f(-x)] \, dx$ with $a=1$. First, we need to find $f(-x)$: $f(-x) = \frac{(1+\sqrt{|-x|-(-x)}) e^{-x}+(\sqrt{|-x|-(-x)}) e^{-(-x)}}{e^{-x}+e^{-(-x)}}$ Since $|-x| = |x|$ and $-(-x) = x$, we have: $f(-x) = \frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{x}}{e^{-x}+e^{x}}$ Now, we compute $f(x) + f(-x)$: $f(x) + f(-x) = \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x}}{e^x+e^{-x}} + \frac{(1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^{x}}{e^x+e^{-x}}$ Combine the numerators over the common denominator: $f(x) + f(-x) = \frac{(1+\sqrt{|x|-x}) e^x+(\sqrt{|x|-x}) e^{-x} + (1+\sqrt{|x|+x}) e^{-x}+(\sqrt{|x|+x}) e^x}{e^x+e^{-x}}$ Group terms with $e^x$ and $e^{-x}$ in the numerator: Numerator $= [(1+\sqrt{|x|-x}) + (\sqrt{|x|+x})] e^x + [(\sqrt{|x|-x}) + (1+\sqrt{|x|+x})] e^{-x}$ Numerator $= (1+\sqrt{|x|-x} + \sqrt{|x|+x}) e^x + (1+\sqrt{|x|-x} + \sqrt{|x|+x}) e^{-x}$ Factor out the common term $(1+\sqrt{|x|-x} + \sqrt{|x|+x})$: Numerator $= (1+\sqrt{|x|-x} + \sqrt{|x|+x}) (e^x + e^{-x})$ So, the sum of the integrands is: $f(x) + f(-x) = \frac{(1+\sqrt{|x|-x} + \sqrt{|x|+x}) (e^x + e^{-x})}{e^x+e^{-x}} = 1+\sqrt{|x|-x} + \sqrt{|x|+x}$ Applying the property, the integral becomes: $I = \int_0^1 (1+\sqrt{|x|-x} + \sqrt{|x|+x}) \, dx$

Step 2: Simplify the Integrand for the Interval $[0, 1]$ For the interval of integration $[0, 1]$, we know that $x \ge 0$. Therefore, $|x| = x$. Substitute $|x|=x$ into the terms involving absolute values:

• $\sqrt{|x|-x} = \sqrt{x-x} = \sqrt{0} = 0$
• $\sqrt{|x|+x} = \sqrt{x+x} = \sqrt{2x}$ The integrand simplifies to: $1 + 0 + \sqrt{2x} = 1 + \sqrt{2x}$ So, the integral is now: $I = \int_0^1 (1+\sqrt{2x}) \, dx$

Step 3: Evaluate the Definite Integral We can rewrite $\sqrt{2x}$ as $\sqrt{2} \cdot \sqrt{x} = \sqrt{2} x^{1/2}$. $I = \int_0^1 (1+\sqrt{2} x^{1/2}) \, dx$ Now, integrate term by term using the power rule: $I = \left[ x + \sqrt{2} \cdot \frac{x^{1/2+1}}{1/2+1} \right]_0^1$ $I = \left[ x + \sqrt{2} \cdot \frac{x^{3/2}}{3/2} \right]_0^1$ $I = \left[ x + \frac{2\sqrt{2}}{3} x^{3/2} \right]_0^1$ Apply the limits of integration (Upper Limit - Lower Limit): $I = \left( 1 + \frac{2\sqrt{2}}{3} (1)^{3/2} \right) - \left( 0 + \frac{2\sqrt{2}}{3} (0)^{3/2} \right)$ $I = \left( 1 + \frac{2\sqrt{2}}{3} \cdot 1 \right) - (0)$ $I = 1 + \frac{2\sqrt{2}}{3}$

Common Mistakes & Tips

• Incorrectly handling absolute values: Ensure you correctly apply the definition of $|x|$ based on the integration interval. In this case, for $[0,1]$, $|x|=x$.
• Algebraic errors during simplification: The simplification of $f(x) + f(-x)$ is crucial. Double-check the factoring and combining of terms to avoid errors.
• Forgetting the symmetry property: The property $\int_{-a}^a f(x) \, dx = \int_0^a [f(x) + f(-x)] \, dx$ is a powerful tool for integrals over symmetric intervals and should be considered as a first step.

Summary The integral was evaluated by first applying the property for definite integrals over symmetric intervals, which transformed the integral from $[-1, 1]$ to $[0, 1]$ and simplified the integrand. Subsequently, the absolute value within the simplified integrand was handled by considering the interval $[0, 1]$, where $|x|=x$. This led to a straightforward integration of the resulting polynomial and power function. The final evaluation using the limits of integration yielded the value $1 + \frac{2\sqrt{2}}{3}$.

The final answer is $\boxed{1+\frac{2 \sqrt{2}}{3}}$.