The value of the integral I = int_{1/2014}^{2 | vedclass

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(B) Let $I = \int_{1/2014}^{2014} \frac{	an^{-1} x}{x} dx$  $(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(\frac{ab}{x}) dx$ is not d

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$\frac{\pi}{2} \log 2014$

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(B) Let $I = \int_{1/2014}^{2014} \frac{	an^{-1} x}{x} dx$  $(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(\frac{ab}{x}) dx$ is not directly applicable, but we use the substitution $x = \frac{1}{t}$, so $dx = -\frac{1}{t^2} dt$.When $x = 1/2014$, $t = 2014$. When $x = 2014$, $t = 1/2014$.$I = \int_{2014}^{1/2014} \frac{	an^{-1}(1/t)}{1/t} (-\frac{1}{t^2}) dt = \int_{1/2014}^{2014} \frac{\cot^{-1} t}{t} dt = \int_{1/2014}^{2014} \frac{\cot^{-1} x}{x} dx$  $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_{1/2014}^{2014} \frac{	an^{-1} x + \cot^{-1} x}{x} dx$Since $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$:$2I = \int_{1/2014}^{2014} \frac{\pi/2}{x} dx = \frac{\pi}{2} [\ln x]_{1/2014}^{2014}$$2I = \frac{\pi}{2} (\ln 2014 - \ln(1/2014)) = \frac{\pi}{2} (\ln 2014 + \ln 2014) = \frac{\pi}{2} (2 \ln 2014) = \pi \ln 2014$$I = \frac{\pi}{2} \log 2014$.

The value of the integral $I = \int_{1/2014}^{2014} \frac{\tan^{-1} x}{x} dx$ is

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