If f(x)=∫1xtan−11tdt, then f(e2)−f1e2=

f(x)=∫1x tan-1tt dt f(1/x)=∫11/x tan-1tt dt Replace t by 1/t =∫1x tan-1(1/t)1/t.-1t2 dt =∫1x -cot-1tt dtf(x)-f1x=∫1x tan-1t+cot-1tt dt =∫1x π2t dt =π2 (log t)1x =π2 (log x) f(e2)-f1e2=π2log e2 =π

If f(x)=∫1xtan−11tdt, then f(e2)−f1e2=

f(x)=∫1x tan-1tt dt f(1/x)=∫11/x tan-1tt dt Replace t by 1/t =∫1x tan-1(1/t)1/t.-1t2 dt =∫1x -cot-1tt dtf(x)-f1x=∫1x tan-1t+cot-1tt dt =∫1x π2t dt =π2 (log t)1x =π2 (log x) f(e2)-f1e2=π2log e2 =π

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If $f (x) = \int_{1}^{x} \frac{\tan^{- 1} 1}{t} d t,$ then $f (e^{2}) - f (\frac{1}{e^{2}}) =$

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$\frac{π}{2}$

$- π$

$\frac{π}{4}$

$π$

answer is D.

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Detailed Solution

$f (x) = \int_{1}^{x} \frac{\tan^{- 1} t}{t} dt f (1 / x) = \int_{1}^{1 / x} \frac{\tan^{- 1} t}{t} dt Replace t by 1 / t = \int_{1}^{x} \frac{\tan^{- 1} (1 / t)}{1 / t} . (\frac{- 1}{t^{2}} dt) = \int_{1}^{x} \frac{- c o t^{- 1} t}{t} dt$

$f (x) - f (\frac{1}{x}) = \int_{1}^{x} \frac{\tan^{- 1} t + \cot^{- 1} t}{t} dt = \int_{1}^{x} \frac{\frac{π}{2}}{t} dt = \frac{π}{2} {(\log t)}_{1}^{x} = \frac{π}{2} (\log x) f (e^{2}) - f (\frac{1}{e^{2}}) = \frac{π}{2} \log e^{2} = π$