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\begin{aligned} f(x) & = \int_\frac 1e^{\tan x} \frac t{1+t^2} dt + \color{#3D99F6}{\int_\frac 1e^{\cot x} \frac 1{t(1+t^2)} dt} \quad \quad \small \color{#3D99F6}{\text{Let }u = \frac 1t \implies du = - \frac 1{t^2} dt \implies \frac {du}{u^2} = - dt} \\ & = \int_\frac 1e^{\tan x} \frac t{1+t^2} dt \color{#3D99F6}{-\int_e^{\tan x} \frac {u\cdot \frac 1{u^2}}{1+\frac 1{u^2}} du} \\ & = \int_\frac 1e^{\tan x} \frac t{1+t^2} dt \color{#D61F06}{+} \int^\color{#D61F06}{e}_\color{#D61F06}{\tan x} \frac u{1+u^2} du \\ & = \int_\frac 1e^e \frac t{1+t^2} dt \\ & = \frac 12 \ln (1+t^2) \bigg|_\frac 1e^e \\ & = \frac 12 \ln \left(\frac {1+e^2}{1+\frac 1{e^2}} \right) \\ & = \frac 12 \ln \left( e^2 \right) \\ & = \boxed{1} \end{aligned}

$f\left( x \right) =\int _{ \frac { 1 }{ e } }^{ \tan { x } }{ \frac { t }{ 1+{ t }^{ 2 } } dt } +\int _{ \frac{1}{e} }^{ cotx }{ \frac { 1 }{ t\left( 1+{ t }^{ 2 } \right) } dt }$ hence using Leibniz integral rule . ( $\because \quad e\in \left[ 0,\frac { \pi }{ 2 } \right]$ ) $\frac { d }{ dx } f\left( x \right) =\frac { \tan { x } }{ 1+{ \tan { x } }^{ 2 } } \left( \frac { d }{ dx } \left\{ \tan { x } \right\} \right) +\frac { 1 }{ \cot { x } \left( 1+{ \cot { x } }^{ 2 } \right) } \left( \frac { d }{ dx } \left\{ \cot { x } \right\} \right)$ We get $f^{ ' }\left( x \right) =\frac { \tan { x } }{ \sec ^{ 2 }{ x } } \sec ^{ 2 }{ x } +\frac { -1 }{ \cot { x } (\csc ^{ 2 }{ x } ) } \csc ^{ 2 }{ x } =0$ Or $f\left( x\right)$ is a constant function.So $f\left( \frac{e}{2}\right)=f\left( \frac { \pi }{ 4 } \right) =\int _{ \frac { 1 }{ e } }^{ 1 }{ \frac { t }{ 1+{ t }^{ 2 } } dt } +\int _{ \frac { 1 }{ e } }^{ 1 }{ \frac { 1 }{ t\left( 1+{ t }^{ 2 } \right) } dt } =\int _{ \frac { 1 }{ e } }^{ e }{ \frac { t }{ 1+{ t }^{ 2 } } dt=\boxed { 1 } } \\$

$f\left( x \right) =\underbrace{\int _{ \frac { 1 }{ e } }^{ \tan { x } }{ \frac { t }{ 1+{ t }^{ 2 } } dt } }_{\mathcal R}+\underbrace{\int _{ \frac { 1 }{ e } }^{ \cot x }{ \frac { 1 }{ t\left( 1+{ t }^{ 2 } \right) } dt }}_{\mathcal I}$

$2\mathcal R= \displaystyle\int _{ \frac { 1 }{ e } }^{ \tan { x } }{ \frac {2 t }{ 1+{ t }^{ 2 } } dt }=\left[\ln(1+t^2)\right]_{1/e}^{\tan x}=\ln(1+\tan^2x)-\ln\left(1+\dfrac1{e^2}\right)$

$-2\mathcal I =\displaystyle\int _{ \frac { 1 }{ e } }^{ \cot x }{ \frac { \frac{-2}{t^3} }{ \left( 1+\frac1{{ t }^{ 2 } }\right) } dt } =\left[\ln\left(1+\frac1{t^2}\right)\right]_{1/e}^{\cot x}=\ln(1+\tan^2 x)-\ln(1+e^2)$

Subtracting and dividing by 2, we get $f(x)=\mathcal R+\mathcal I=\dfrac 12\ln e^2=1$

Hence, $\large f\left(\frac{e}{2}\right) =\boxed 1$

$\int _{ 1/e }^{ tan(x) }{ \frac { t }{ 1+{ t }^{ 2 } } } dt\quad =\quad A\\ let\quad 1+{ t }^{ 2 }=y,\quad then\quad t.dt=dy/2\\ A=\frac { 1 }{ 2 } \int _{ 1/e }^{ tan(x) }{ \frac { dy }{ y } } =\quad \frac { 1 }{ 2 } ln(\frac { \sec ^{ 2 }{ x } { e }^{ 2 } }{ { e }^{ 2 }+1 } )\\ Also\\ \int _{ 1/e }^{ cot(x) }{ \frac { dt }{ t(1+{ t }^{ 2 }) } } =\quad B\quad =\quad \int _{ 1/e }^{ cot(x) }{ \frac { t.dt }{ { t }^{ 2 }(1+{ t }^{ 2 }) } } \\ let\quad 1+{ t }^{ 2 }=y,\quad then\quad t.dt=dy/2\\ B\quad =\quad \frac { 1 }{ 2 } \int _{ 1/e }^{ cot(x) }{ \frac { dy }{ (y-1)y } } \\ Apply\quad partial\quad fractions\quad technique,\quad Assume\quad two\quad constants\quad C\quad and\quad D\quad such\quad that\\ B\quad =\quad \frac { C }{ y } +\frac { D }{ y-1 } \quad (2)\\ B\quad =\quad \frac { 1 }{ 2 } (C\int _{ 1/e }^{ cot(x) }{ \frac { dy }{ y } } +\quad D\int _{ 1/e }^{ cot(x) }{ \frac { dy }{ y-1 } } )\\ B\quad =\quad \frac { 1 }{ 2 } (C.ln(\frac { \csc ^{ 2 }{ (x).{ e }^{ 2 } } }{ { e }^{ 2 }+1 } )\quad +\quad D.ln(\cot ^{ 2 }{ (x).{ e }^{ 2 } } ))\\ from\quad (1),\quad C(y-1)\quad +\quad Dy\quad =\quad 1\quad (equating\quad numerators)\\ we\quad get\quad C=-D\quad C=-1,\quad hence\quad C=-1,D=1\\ we\quad get,\\ B\quad =\quad \frac { 1 }{ 2 } ln(\cos ^{ 2 }{ (x).{ e }^{ 2 } } )\\ hence,\quad f(x)\quad =\quad A\quad +\quad B\quad =\quad \frac { 1 }{ 2 } \quad ln({ e }^{ 2 })\quad =\quad 1,\quad hence\quad f(x)\quad is\quad a\quad constant\quad function\\ \boxed { f(\frac { e }{ 2 } )\quad =\quad 1 }$

$f(x) =\displaystyle\int_{\frac{1}{e}}^{\tan x}\dfrac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\dfrac{1}{t(1+t^2)}dt$

First, split the fraction on the right integral into partial fractions

$\dfrac{1}{t(1+t^2)}=\dfrac{a}{t} + \dfrac{bt+c}{1+t^2}\\ a(1+t^2)+t(bt+c)=1\\ (a+b)t^2+ct+a=1\\ a=1,\;c=0,\;a+b=0\implies b=-a=-1\\ \implies\dfrac{1}{t(1+t^2)}=\dfrac{1}{t} - \dfrac{t}{1+t^2}$

Therefore,

$f(x) = \displaystyle\int_{\frac{1}{e}}^{\tan x}\dfrac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\left(\dfrac{1}{t} - \dfrac{t}{1+t^2}\right)dt\\ = \displaystyle\int_{\frac{1}{e}}^{\tan x}\dfrac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\dfrac{1}{t} dt-\int_{\frac{1}{e}}^{\cot x} \dfrac{t}{1+t^2}dt\\ =\displaystyle\int_{\frac{1}{e}}^{\tan x}\dfrac{t}{1+t^2}dt\color{#3D99F6}{+\int_{\cot x}^{\frac{1}{e}}} \dfrac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\dfrac{1}{t} dt\\ =\color{#D61F06}{\dfrac{1}{2}}\displaystyle\color{#3D99F6}{\int_{\cot x}^{\tan x}} \dfrac{\color{#D61F06}{2}t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\dfrac{1}{t} dt\\ =\dfrac{1}{2}\Big[\ln(1+t^2)\Big]_{\cot x}^{\tan x} + \Big[\ln x\Big]_{\frac{1}{e}}^{\cot x}\\ =\dfrac{1}{2}\Big(\ln(\color{#20A900}{1+\tan^2 x})-\ln(\color{#EC7300}{1+\cot^2 x})\Big)+\Big(\ln(\cot x)-\ln\left(\dfrac{1}{e}\right)\Big)\\ =\dfrac{1}{2}\Big(\ln(\color{#20A900}{\sec^2 x})-\ln(\color{#EC7300}{\csc^2 x})\Big)+\Big(\ln(\cot x)-\ln\left(e^{-1}\right)\Big)\\ =\dfrac{1}{2}\Big(\ln\left(\dfrac{\sec^2 x}{\csc^2 x}\right)\Big)+\Big(\ln(\cot x)+1\Big)\\ =\dfrac{1}{2}\Big(\ln\left(\dfrac{\sin x}{\cos x}\right)^2\Big)+\ln(\cot x)+1\\ =\ln(\tan x)+\ln\left(\dfrac{1}{\tan x}\right)+1\\ =\ln\left(\tan x\left(\dfrac{1}{\tan x}\right)\right)+1\\ =\ln1 + 1\\ =0+1\\ =1$

$\implies f\left(\dfrac{e}{2}\right)=\boxed{1}$

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Trigonometric identities used :

1. $\color{#20A900}{1+\tan^2 x = \sec^2x}$
2. $\color{#EC7300}{1+\cot^2 x = \csc^2 x}$

The integrands differ only in the placement of $t$ (numerator vs. denominator). This gave me the idea to rewrite the second integral with $u = 1/t$ . Thus: $\frac{1}{t(1+t^2)} dt = \frac{u}{1 + (\tfrac 1u)^2} \frac {-du}{u^2} = -\frac{u}{1 + u^2} du.$ Thus the second integral becomes (accounting for the negative sign by flipping the boundaries) $\int_{\frac 1 e}^{\cot x} \frac 1{t(1 + t^2)} dt = \int_{\tan x}^{e} \frac u {1 + u^2} du,$ and at this point it is clear that the two integrals can be combined into one. Replacing $u$ by $t$ in our last equation, we simply get $f(x) = \int_{\frac 1 e}^{\tan x} \frac t {1 + t^2} dt + \int_{\tan x}^{e} \frac t {1 + t^2} dt = \int_{\frac 1 e}^{e} \frac t {1 + t^2} dt.$ The variable $x$ no longer occurs, so this is a constant function. The integral is easily solved (noting that the numerator is the derivative of the denominator) $f(x) = \int_{\frac 1 e}^{e} \frac t {1 + t^2} dt = \left.\frac 1 2 \ln (1 + t^2)\right|_{1/e}^e = \frac 12\left(\ln (1 + e^2) - \ln (1 + e^{-2})\right) \\ = \frac12 \ln\left(\frac{1 + e^2}{1 + e^{-2}}\right) = \frac12 \ln e^2 = \boxed{1}.$