1. Key Concepts and Formulas

• Substitution Rule for Definite Integrals: If $x = g(u)$ and $dx = g'(u)du$, then $\int_a^b f(x)dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(u))g'(u)du$. The limits of integration must be transformed according to the substitution.
• Properties of Definite Integrals: $\int_a^b f(x)dx = -\int_b^a f(x)dx$.
• Inverse Trigonometric Identity: For $x > 0$, $\tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$.
• Logarithm Properties: $\ln(a/b) = \ln a - \ln b$, $\ln(1/a) = -\ln a$, and $\ln(a^b) = b\ln a$.

The limits of integration, $1/2$ and $2$, are reciprocals of each other. This strongly suggests a substitution of the form $x = 1/u$.

2. Step-by-Step Solution

Step 1: Define the Integral and Identify the Strategy Let the given integral be $I$. $I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} \quad \ldots (1)$ The limits of integration are reciprocals ($1/2$ and $2$). This indicates that a substitution $x = 1/u$ will be beneficial to transform the limits and the integrand.

Step 2: Apply the Substitution Let $x = \frac{1}{u}$. Then, $dx = -\frac{1}{u^2}du$.

We must also change the limits of integration:

• When $x = \frac{1}{2}$, $u = \frac{1}{x} = \frac{1}{1/2} = 2$.
• When $x = 2$, $u = \frac{1}{x} = \frac{1}{2}$.

Step 3: Transform the Integral Substitute $x = 1/u$, $dx = -1/u^2 du$, and the new limits into the integral $I$: $I = \int\limits_{2}^{1/2} {{{\tan^{-1}\left(\frac{1}{u}\right)} \over {\left(\frac{1}{u}\right)}} \left(-\frac{1}{u^2}\right)du}$ Simplify the integrand: $I = \int\limits_{2}^{1/2} {\frac{\tan^{-1}\left(\frac{1}{u}\right)}{1} \cdot u \cdot \left(-\frac{1}{u^2}\right)du}$ $I = \int\limits_{2}^{1/2} {- \frac{\tan^{-1}\left(\frac{1}{u}\right)}{u} du}$ Using the property $\int_a^b f(u)du = -\int_b^a f(u)du$, we can swap the limits and remove the negative sign: $I = \int\limits_{1/2}^{2} {\frac{\tan^{-1}\left(\frac{1}{u}\right)}{u} du}$ Since $u$ is a dummy variable, we can replace it with $x$: $I = \int\limits_{1/2}^{2} {\frac{\tan^{-1}\left(\frac{1}{x}\right)}{x} dx} \quad \ldots (2)$

Step 4: Combine the Original and Transformed Integrals We now have two expressions for $I$: (1) $I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx}$ (2) $I = \int\limits_{{1 \over 2}}^2 {{{\tan^{-1}\left(\frac{1}{x}\right)} \over x}dx}$

Adding (1) and (2): $I + I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} + \int\limits_{{1 \over 2}}^2 {{{\tan^{-1}\left(\frac{1}{x}\right)} \over x}dx}$ Since the limits of integration are the same, we can combine the integrands: $2I = \int\limits_{{1 \over 2}}^2 {\left( {{{{{\tan }^{ - 1}}x} \over x} + {{\tan^{-1}\left(\frac{1}{x}\right)} \over x}} \right)dx}$ 2I = \int\limits_{{1 \over 2}}^2 {\frac{1}{x}\left( {{{\tan }^{ - 1}}x + {{\tan^{-1}\left(\frac{1}{x}\right)}} \right)dx}

Step 5: Apply the Inverse Trigonometric Identity For $x > 0$, we know that $\tan^{-1}x + \tan^{-1}\left(\frac{1}{x}\right) = \frac{\pi}{2}$. Since the limits of integration are $1/2$ and $2$, $x$ is always positive in this interval. Thus, we can apply the identity: $2I = \int\limits_{{1 \over 2}}^2 {\frac{1}{x}\left( {\frac{\pi }{2}} \right)dx}$ $2I = \frac{\pi }{2} \int\limits_{{1 \over 2}}^2 {\frac{1}{x}dx}$

Step 6: Evaluate the Simplified Integral The integral of $1/x$ is $\ln|x|$: $2I = \frac{\pi }{2} \left[ {\ln|x|} \right]_{{1 \over 2}}^2$ Apply the limits of integration: $2I = \frac{\pi }{2} \left( {\ln|2| - \ln\left|\frac{1}{2}\right|} \right)$ Since $2 > 0$ and $1/2 > 0$: $2I = \frac{\pi }{2} \left( {\ln 2 - \ln\left(\frac{1}{2}\right)} \right)$ Using the logarithm property $\ln(1/a) = -\ln a$: $2I = \frac{\pi }{2} \left( {\ln 2 - (-\ln 2)} \right)$ $2I = \frac{\pi }{2} \left( {\ln 2 + \ln 2} \right)$ $2I = \frac{\pi }{2} (2 \ln 2)$ $2I = \pi \ln 2$

Step 7: Solve for I Divide by 2 to find the value of $I$: $I = \frac{\pi }{2} \ln 2$

3. Common Mistakes & Tips

• Forgetting to Change Limits: When performing a substitution in a definite integral, always remember to change the limits of integration to correspond to the new variable. Forgetting this is a very common error.
• Incorrect Inverse Tangent Identity Application: The identity $\tan^{-1}x + \tan^{-1}(1/x) = \pi/2$ is valid only for $x > 0$. Ensure the domain of integration satisfies this condition.
• Algebraic Errors with Logarithms: Be careful with logarithm properties, especially when dealing with fractions or negative exponents, such as $\ln(1/2)$.

4. Summary

The integral was evaluated by recognizing that the limits of integration were reciprocals. A substitution $x = 1/u$ was applied, which transformed the integral and its limits. By adding the original integral to the transformed integral, and using the identity $\tan^{-1}x + \tan^{-1}(1/x) = \pi/2$, the integrand was simplified to a constant multiple of $1/x$. This allowed for a straightforward evaluation of the integral, resulting in the value $\frac{\pi}{2} \log_e 2$.

The final answer is $\boxed{{\pi \over 2}{\log _e}2}$.