Key Concepts and Formulas

• Algebraic Identity: $a^2 + b^2 - 2ab = (a-b)^2$.
• Property of Square Roots: $\sqrt{X^2} = |X|$.
• Properties of Definite Integrals: For an even function $f(x)$, $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$.
• Integral of $\frac{g'(x)}{g(x)}$: $\int \frac{g'(x)}{g(x)} dx = \ln|g(x)| + C$.
• Logarithm Properties: $\ln(a^b) = b \ln(a)$, $\ln(1/a) = -\ln(a)$.

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Step-by-Step Solution

Let the given integral be $I$. $I = \int\limits_{{{ - 1} \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left( {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right)}^{{1 \over 2}}}dx}$

Step 1: Simplify the Integrand We focus on the expression inside the square root: ${\left( {{{x + 1} \over {x - 1}}} \right)^2} + {\left( {{{x - 1} \over {x + 1}}} \right)^2} - 2$ This expression is in the form $a^2 + b^2 - 2$, where $a = \frac{x+1}{x-1}$ and $b = \frac{x-1}{x+1}$. Since $ab = \left(\frac{x+1}{x-1}\right)\left(\frac{x-1}{x+1}\right) = 1$, the expression is $a^2 + b^2 - 2ab$. Using the algebraic identity $a^2 + b^2 - 2ab = (a-b)^2$, we get: ${\left( {\left( {{{x + 1} \over {x - 1}}} \right) - \left( {{{x - 1} \over {x + 1}}} \right)} \right)^2}$ Now, we take the square root of this expression: $\sqrt{{{\left( {\left( {{{x + 1} \over {x - 1}}} \right) - \left( {{{x - 1} \over {x + 1}}} \right)} \right)^2}}} = \left| {\left( {{{x + 1} \over {x - 1}}} \right) - \left( {{{x - 1} \over {x + 1}}} \right)} \right|$ This step is crucial because $\sqrt{X^2} = |X|$, not simply $X$.

Next, we simplify the expression inside the absolute value by finding a common denominator: $\left| {\frac{{(x + 1)(x + 1) - (x - 1)(x - 1)}}{{(x - 1)(x + 1)}}} \right| = \left| {\frac{{{{(x + 1)}^2} - {{(x - 1)}^2}}}{{{x^2} - 1}}} \right|$ Expanding the numerator using $(A+B)^2 - (A-B)^2 = 4AB$: $\left| {\frac{{({x^2} + 2x + 1) - ({x^2} - 2x + 1)}}{{{x^2} - 1}}} \right| = \left| {\frac{{4x}}{{{x^2} - 1}}} \right|$ So, the integral becomes: $I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}$

Step 2: Analyze the Integrand for Symmetry The limits of integration are from $-\frac{1}{\sqrt{2}}$ to $\frac{1}{\sqrt{2}}$, which are symmetric about 0. Let $f(x) = \left| \frac{4x}{x^2 - 1} \right|$. We check if $f(x)$ is an even or odd function: $f(-x) = \left| \frac{4(-x)}{(-x)^2 - 1} \right| = \left| \frac{-4x}{x^2 - 1} \right|$ Since $|-y| = |y|$, we have: $f(-x) = \left| \frac{4x}{x^2 - 1} \right| = f(x)$ Thus, $f(x)$ is an even function. This allows us to use the property $\int_{-a}^a f(x) dx = 2 \int_0^a f(x) dx$. $I = 2 \int\limits_{0}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}$

Step 3: Evaluate the Absolute Value within the Integration Interval For the interval $0 \le x < \frac{1}{\sqrt{2}}$: The numerator $4x$ is non-negative. The denominator $x^2 - 1$: Since $0 \le x < \frac{1}{\sqrt{2}}$, we have $0 \le x^2 < \frac{1}{2}$. Therefore, $x^2 - 1$ is negative. So, $\frac{4x}{x^2 - 1}$ is negative. $\left| {{{4x} \over {{x^2} - 1}}} \right| = - \left( {{{4x} \over {{x^2} - 1}}} \right) = \frac{{4x}}{{1 - {x^2}}}$ The integral becomes: $I = 2 \int\limits_{0}^{{1 \over {\sqrt 2 }}} {\frac{{4x}}{{1 - {x^2}}}dx}$

Step 4: Evaluate the Definite Integral We can rewrite the integral as: $I = 8 \int\limits_{0}^{{1 \over {\sqrt 2 }}} {\frac{x}{{1 - {x^2}}}dx}$ To integrate $\frac{x}{1-x^2}$, we can use substitution or recognize it as a form of $\frac{g'(x)}{g(x)}$. Let $u = 1 - x^2$. Then $du = -2x dx$, which means $x dx = -\frac{1}{2} du$. When $x=0$, $u = 1 - 0^2 = 1$. When $x = \frac{1}{\sqrt{2}}$, $u = 1 - \left(\frac{1}{\sqrt{2}}\right)^2 = 1 - \frac{1}{2} = \frac{1}{2}$. The integral transforms to: $I = 8 \int\limits_{1}^{{1 \over 2}} {\frac{1}{u} \left( {-\frac{1}{2}du} \right)} = 8 \left(-\frac{1}{2}\right) \int\limits_{1}^{{1 \over 2}} {\frac{1}{u}du}$ $I = -4 \int\limits_{1}^{{1 \over 2}} {\frac{1}{u}du}$ Now, we evaluate the integral: $I = -4 \left[ {\ln|u|} \right]_{1}^{{1 \over 2}} = -4 \left( {\ln\left(\frac{1}{2}\right) - \ln(1)} \right)$ Since $\ln(1) = 0$: $I = -4 \left( {\ln\left(\frac{1}{2}\right)} \right)$ Using the logarithm property $\ln(1/a) = -\ln(a)$: $I = -4 \left( {-\ln(2)} \right) = 4 \ln(2)$ Using the logarithm property $b \ln(a) = \ln(a^b)$: $I = \ln(2^4) = \ln(16)$

Common Mistakes & Tips

• Forgetting the Absolute Value: The square root of a squared term is always the absolute value. Failing to include this can lead to incorrect signs in the integration.
• Incorrectly Handling Absolute Value: Carefully analyze the sign of the expression inside the absolute value over the given interval.
• Symmetry Property Application: Ensure the function is indeed even or odd before applying the symmetry properties for definite integrals.

Summary

The problem was solved by first simplifying the integrand using algebraic identities. The resulting expression involved an absolute value, which was then handled by analyzing the sign of the function within the integration interval. The symmetric limits of integration allowed us to use the property of even functions, reducing the problem to integrating over half the interval. Finally, a standard logarithmic integral was evaluated to obtain the answer.

The final answer is $\boxed{\ln 16}$.