Finding value of given integral Find the value of the following integral. $$\int_{1/e}^{\tan x} \cfrac{tdt}{1+t^2} + \int_{1/e}^{\cot x} \cfrac{dt}{1+t^2}$$ 
As a start, I've done this - 
$$\int_{1/e}^{\tan x} \cfrac{tdt}{1+t^2} - \int_{\cot x}^{1/e} \cfrac{dt}{1+t^2} \\
=\int_{\cot x}^{\tan x} \cfrac{t-1}{1+t^2} dt $$
Is this valid? If yes, then how to proceed?
 A: The integrands are easy to... well, to integrate. The primitive of the first is $\frac12\ln(1+t^2)$ and of the second, $\arctan t$. Apply the Barrow's rule to get:
$$\frac12[\ln(1+\tan^2 x)+1]+\arctan\cot x -\arctan\left(\frac1e\right)$$
A: $$\int_{1/e}^{\tan x} \cfrac{tdt}{1+t^2} + \int_{1/e}^{\cot x} \cfrac{dt}{1+t^2}$$
For the first integral, substitute $1+t^2=s\implies tdt= \dfrac {ds}2$. Also, we know that $\int\frac 1{1+t^2} dt =\arctan t +c$. We thus get: $$\int_{1/e}^{\tan x} \dfrac{ds}{2s} + \arctan (\cot x)-\arctan\left (\dfrac 1e\right )$$ $$=\dfrac 12(\ln \tan x +1)+\arctan (\cot x)-\arctan\left (\dfrac 1e\right )$$
A: $$\int _{ 1/e }^{ tanx }{ \frac { tdt }{ 1+{ t }^{ 2 } } = } \frac { 1 }{ 2 } \int _{ 1/e }^{ tanx }{ \frac { d\left( 1+{ t }^{ 2 } \right)  }{ 1+{ t }^{ 2 } } = } \frac { 1 }{ 2 } ln\left( 1+{ t }^{ 2 } \right) |_{ 1/e }^{ tanx }=\frac { 1 }{ 2 } \left[ ln\frac { \sec ^{ 2 }{ x }  }{ 1+{ e }^{ -2 } }  \right] +C$$
$$ \int _{ 1/e }^{ cotx }{ \frac { dt }{ 1+{ t }^{ 2 } } =\arctan { t|_{ 1/e }^{ cotx } } = } \arctan { \cot { x } -\arctan { 1/e }  } +C$$
