In the second integral, we can factor out (√(|x|) - x) from the numerator, which simplifies the expression to (√(|x|) - x) (e^x + e^(-x))/(e^x + e^(-x)) = (√(|x|) - x).

We can further split the second integral into two separate integrals: one for √(|x|) and one for x. This allows us to evaluate each part individually.

For an integral from -a to a: if f(x) is an odd function (f(-x) = -f(x)), then ∫_-a^(a) f(x) dx = 0. Since x is an odd function, ∫_-1^(1) x dx = 0. If f(x) is an even function (f(-x) = f(x)), then ∫_-a^(a) f(x) dx = 2 ∫_0^(a) f(x) dx. Since √(|x|) is an even function, ∫_-1^(1) √(|x|) dx = 2 ∫_0^(1) √x dx.

For an integral from -a to a: if f(x) is an odd function (f(-x) = -f(x)), then ∫_-a^(a) f(x) dx = 0. Since x is an odd function, ∫_-1^(1) x dx = 0. If f(x) is an even function (f(-x) = f(x)), then ∫_-a^(a) f(x) dx = 2 ∫_0^(a) f(x) dx. Since √(|x|) is an even function, ∫_-1^(1) √(|x|) dx = 2 ∫_0^(1) √x dx.

For the first integral, multiply the numerator and denominator by e^x. Let u = e²x + 1, then du = 2e²x dx. The integral becomes (1)/(2) ∫ (1)/(u) du = (1)/(2) ln|u|. Evaluating from -1 to 1 gives (1)/(2) [ln(e²x + 1)]_-1^(1) = (1)/(2) (ln(e²+1) - ln(e^(-2)+1)). For the second integral, we evaluate ∫ x^(1/2) dx = x³/23/2 = (2)/(3) x³/2 from 0 to 1.

For the first integral, multiply the numerator and denominator by e^x. Let u = e²x + 1, then du = 2e²x dx. The integral becomes (1)/(2) ∫ (1)/(u) du = (1)/(2) ln|u|. Evaluating from -1 to 1 gives (1)/(2) [ln(e²x + 1)]_-1^(1) = (1)/(2) (ln(e²+1) - ln(e^(-2)+1)). For the second integral, we evaluate ∫ x^(1/2) dx = x³/23/2 = (2)/(3) x³/2 from 0 to 1.

The value of $\int_{-1}^{1} \frac{(1 + \sqrt{|x|} - x)e^x + (\sqrt{|x|} - x)e^{-x}}{e^x + e^{-x}}$ \, dx is equal to:

Answer: $\frac{7}{3}$

Step-by-step solution

Step 1: Split the integrand

First, we can split the given integral into two parts. The first part contains the $e^x$ term from the numerator, and the second part contains the remaining terms. This separation helps simplify the expression.

Step 2: Simplify the second integral

In the second integral, we can factor out $(\sqrt{|x|} - x)$ from the numerator, which simplifies the expression to $(\sqrt{|x|} - x) \frac{e^x + e^{-x}}{e^x + e^{-x}} = (\sqrt{|x|} - x)$ .

Step 3: Split the second integral further

We can further split the second integral into two separate integrals: one for $\sqrt{|x|}$ and one for $x$ . This allows us to evaluate each part individually.

Step 4: Evaluate integrals of odd and even functions

For an integral from $-a$ to $a$ : if $f(x)$ is an odd function ( $f(-x) = -f(x)$ ), then $\int_{-a}^{a} f(x) \, dx = 0$ . Since $x$ is an odd function, $\int_{-1}^{1} x \, dx = 0$ . If $f(x)$ is an even function ( $f(-x) = f(x)$ ), then $\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$ . Since $\sqrt{|x|}$ is an even function, $\int_{-1}^{1} \sqrt{|x|} \, dx = 2 \int_{0}^{1} \sqrt{x} \, dx$ .

Step 5: Evaluate the remaining integrals

For the first integral, multiply the numerator and denominator by $e^x$ . Let $u = e^{2x} + 1$ , then $du = 2e^{2x} \, dx$ . The integral becomes $\frac{1}{2} \int \frac{1}{u} \, du = \frac{1}{2} \ln|u|$ . Evaluating from $-1$ to $1$ gives $\frac{1}{2} [\ln(e^{2x} + 1)]_{-1}^{1} = \frac{1}{2} (\ln(e^2+1) - \ln(e^{-2}+1))$ . For the second integral, we evaluate $\int x^{1/2} \, dx = \frac{x^{3/2}}{3/2} = \frac{2}{3} x^{3/2}$ from $0$ to $1$ .

Step 6: Combine the results

Substitute the evaluated values back into the expression for $I$ . The logarithmic terms simplify to $1$ . The term $2 \times \frac{2}{3}$ becomes $\frac{4}{3}$ .

Step 7: Final Calculation

Add the simplified terms to get the final value of the integral.

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The value of ∫−11(1+∣x∣−x)ex+(∣x∣−x)e−xex+e−x\int_{-1}^{1} \frac{(1 + \sqrt{|x|} - x)e^x + (\sqrt{|x|} - x)e^{-x}}{e^x + e^{-x}}∫−11​ex+e−x(1+∣x∣​−x)ex+(∣x∣​−x)e−x​ \, dx is equal to: