The value of $\int\limits_{1/e}^{tan\,x}\frac{tdt}{1+t^2}+\int\limits_{1/e}^{cot\,x}\frac{dt}{t(1+t^2)}$ is equal to

1

Let $I(x)=\int\limits_{1/e}^{tan\,x}\frac{tdt}{1+t^2}+\int\limits_{1/e}^{cot\,x}\frac{dt}{t(1+t^2)}=\int\limits_{1/e}^{tan\,x}\frac{tdt}{1+t^2}+\int\limits_{1/e}^{cot\,x}(\frac{1}{t}-\frac{t}{1+t^2})dt$

$=\int\limits_{cot\,x}^{tan\,x}\frac{tdt}{1+t^2}+\int\limits_{1/e}^{cot\,x}\frac{1}{t}dt=\frac{1}{2}In(1+t^2)|_{cot\,x}^{tan\,x}+In\,t|_{1/e}^{cot\,x}$

= – ln cot x + ln cot x – ln $\frac{1}{e}$ = 1

Hence (A) is the correct answer.