The value of the integral int limits_{1 / 2}^ | vedclass

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(D) Let $I = \int \limits_{1 / 2}^2 \frac{	an ^{-1} x}{x} dx$  ... $(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(\frac{ab}{x}) dx$,we substi

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$\frac{\pi}{2} \log _{ e } 2$

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(D) Let $I = \int \limits_{1 / 2}^2 \frac{	an ^{-1} x}{x} dx$  ... $(i)$Using the property $\int_a^b f(x) dx = \int_a^b f(\frac{ab}{x}) dx$,we substitute $x = \frac{1}{t}$,so $dx = -\frac{1}{t^2} dt$.When $x = 1/2, t = 2$ and when $x = 2, t = 1/2$.$I = \int \limits_2^{1 / 2} \frac{	an ^{-1}(1/t)}{1/t} \cdot (-\frac{1}{t^2}) dt = \int \limits_{1 / 2}^2 \frac{	an ^{-1}(1/t)}{t} dt$.Since $	an^{-1}(1/t) = \cot^{-1} t$ for $t > 0$,we have $I = \int \limits_{1 / 2}^2 \frac{\cot ^{-1} t}{t} dt = \int \limits_{1 / 2}^2 \frac{\cot ^{-1} x}{x} dx$ ... (ii)Adding $(i)$ and (ii):$2I = \int \limits_{1 / 2}^2 \frac{	an ^{-1} x + \cot ^{-1} x}{x} dx = \int \limits_{1 / 2}^2 \frac{\pi / 2}{x} dx$.$2I = \frac{\pi}{2} [\ln x]_{1/2}^2 = \frac{\pi}{2} (\ln 2 - \ln(1/2)) = \frac{\pi}{2} (\ln 2 - (-\ln 2)) = \frac{\pi}{2} (2 \ln 2) = \pi \ln 2$.Therefore,$I = \frac{\pi}{2} \ln 2$.

The value of the integral $\int \limits_{1 / 2}^2 \frac{\tan ^{-1} x}{x} d x$ is equal to

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