The value of the integral int_{-1}^{1} frac{d | vedclass

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(C) Let $f(x) = 	an^{-1} \frac{1}{x}$. The integral is $\int_{-1}^{1} f'(x) dx = [f(x)]_{-1}^{1} = f(1) - f(-1)$.Since $f(1) = 	an^{-1}(1) = \frac{\

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$-\frac{\pi}{2}$

Vedclass · Answer

(C) Let $f(x) = 	an^{-1} \frac{1}{x}$. The integral is $\int_{-1}^{1} f'(x) dx = [f(x)]_{-1}^{1} = f(1) - f(-1)$.Since $f(1) = 	an^{-1}(1) = \frac{\pi}{4}$ and $f(-1) = 	an^{-1}(-1) = -\frac{\pi}{4}$.Therefore,$\int_{-1}^{1} f'(x) dx = \frac{\pi}{4} - (-\frac{\pi}{4}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$.However,note that the function $	an^{-1} \frac{1}{x}$ is discontinuous at $x = 0$. The integral is an improper integral: $\int_{-1}^{1} f'(x) dx = \lim_{\epsilon 	o 0^+} \int_{-1}^{-\epsilon} f'(x) dx + \lim_{\delta 	o 0^+} \int_{\delta}^{1} f'(x) dx$.$= [f(-\epsilon) - f(-1)] + [f(1) - f(\delta)] = (	an^{-1}(-1/\epsilon) - 	an^{-1}(-1)) + (	an^{-1}(1) - 	an^{-1}(1/\delta))$.As $\epsilon, \delta 	o 0$,$	an^{-1}(-1/\epsilon) 	o -\frac{\pi}{2}$ and $	an^{-1}(1/\delta) 	o \frac{\pi}{2}$.Result $= (-\frac{\pi}{2} - (-\frac{\pi}{4})) + (\frac{\pi}{4} - \frac{\pi}{2}) = (-\frac{\pi}{2} + \frac{\pi}{4}) + (\frac{\pi}{4} - \frac{\pi}{2}) = -\frac{\pi}{4} - \frac{\pi}{4} = -\frac{\pi}{2}$.

The value of the integral $\int_{-1}^{1} \frac{d}{dx} \left( \tan^{-1} \frac{1}{x} \right) dx$ is

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