<s>[INST] Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$ 
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$

$\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$
Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get
$$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$
So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$
So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$
My question is how can we solve it in some shorter way, Help required, Thanks
 [/INST] Just differentiate the $f(x)$ wrt $x$, you will se that $f'(x)=0$ which means $f(x)=constant=p$. SO you can find $p$ by just putting $x=\frac{\pi}{4}$. So 
$$p=\int_{\frac{1}{e}}^{1} \frac{tdt}{t^2+1}+\int_{\frac{1}{e}}^{1} \frac{dt}{t(t^2+1)}=\int_{\frac{1}{e}}^{1}\frac{dt}{t}=0-(-1)=1$$
SO $f(x)=1$ which is the final answer
Hope this will be helpful !
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