Let f(x) = intlimits_1^x frac{tan^{-1} t}{t} | vedclass

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(B) We are given $f(x) = \int_{1}^{x} \frac{	an^{-1} t}{t} dt$. We need to evaluate $f(e^2) - f\left(\frac{1}{e^2}ight) = \int_{1}^{e^2} \frac{	an

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$\pi$

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(B) We are given $f(x) = \int_{1}^{x} \frac{	an^{-1} t}{t} dt$. We need to evaluate $f(e^2) - f\left(\frac{1}{e^2}ight) = \int_{1}^{e^2} \frac{	an^{-1} t}{t} dt - \int_{1}^{1/e^2} \frac{	an^{-1} t}{t} dt$. Note that $-\int_{1}^{1/e^2} \frac{	an^{-1} t}{t} dt = \int_{1/e^2}^{1} \frac{	an^{-1} t}{t} dt$. So,$f(e^2) - f\left(\frac{1}{e^2}ight) = \int_{1}^{e^2} \frac{	an^{-1} t}{t} dt + \int_{1/e^2}^{1} \frac{	an^{-1} t}{t} dt = \int_{1/e^2}^{e^2} \frac{	an^{-1} t}{t} dt$. Let $I = \int_{1/e^2}^{e^2} \frac{	an^{-1} t}{t} dt$. Using the property $\int_{a}^{b} g(t) dt = \int_{a}^{b} g\left(\frac{ab}{t}ight) dt$,where $ab = (e^2)(1/e^2) = 1$,we have $I = \int_{1/e^2}^{e^2} \frac{	an^{-1}(1/t)}{t} dt$. Adding the two expressions for $I$: $2I = \int_{1/e^2}^{e^2} \frac{	an^{-1} t + 	an^{-1}(1/t)}{t} dt$. Since $	an^{-1} t + 	an^{-1}(1/t) = \frac{\pi}{2}$ for $t > 0$,we get $2I = \int_{1/e^2}^{e^2} \frac{\pi/2}{t} dt = \frac{\pi}{2} [\ln t]_{1/e^2}^{e^2} = \frac{\pi}{2} (\ln e^2 - \ln e^{-2}) = \frac{\pi}{2} (2 - (-2)) = \frac{\pi}{2} (4) = 2\pi$. Therefore,$I = \frac{2\pi}{2} = \pi$.

Let $f(x) = \int\limits_1^x \frac{\tan^{-1} t}{t} dt$ for $x > 0$. Then $f(e^2) - f\left(\frac{1}{e^2}\right)$ is

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