An interesting integral $I = \int\limits_{-1}^{1} \arctan(e^x)dx $ I solved this interesting integral online:
$$I  = \int\limits_{-1}^{1} \arctan(e^x)dx $$
Now I tried the substitution $u=e^x$ but it lead me nowhere.  I was looking at the following post which was solved in a beautiful way Integrate $\int_0^{\pi/2} \frac{1}{1+\tan^\alpha{x}}\,\mathrm{d}x$. From there I found this very interesting article http://www.maa.org/sites/default/files/pdf/mathdl/CMJ/Nelsen39-41.pdf which has the integral I posted at the end as a question to the reader. 
Looking at the graph of $\arctan(e^x)dx$ on the interval $-1 \leq x \leq 1$ I conjectured that $I=\frac{\pi}{2}$. I used the following method to prove it:
\begin{eqnarray}
-e^{-x}&=&\frac{-1}{e^x}\\&=& \frac{-1}{\tan\{   \arctan(e^x)  \} }\\&=&-\cot\{ \arctan(e^x) \}\\
&=& \tan \left\{\arctan(e^x)-\frac{\pi}{2} \right\}\\
\end{eqnarray}
For the last equality I used the fact that $\cot(\theta) =-\tan\left(\theta -\frac{\pi}{2}\right) $. Now we take the arctan of both sides to obtain:
$$\arctan(-e^{-x}) = \arctan(e^x)-\frac{\pi}{2}$$
Finally I use the fact that $\arctan(-\theta)=-\arctan(\theta)$ and add $\frac{\pi}{4}$ to both sides of the last equation to obtain:
$$-\arctan(e^{-x})   +\frac{\pi}{4}= \arctan(e^x)-\frac{\pi}{4} $$
So it is established that the function $f(x) = \arctan(e^x)-\frac{\pi}{4}$ is an odd function. Thus
$$I_2  = \int\limits_{-1}^{1}  \left[ \arctan(e^x)-\frac{\pi}{4} \right]dx = 0 $$
Now $$I =  \int\limits_{-1}^{1} \arctan(e^x)dx  = I_2 +  \int\limits_{-1}^{1} \frac{\pi}{4}dx =\frac{\pi}{2} $$
I thought this integral was really interesting and I was wondering if anyone else has any different ways of solving it, possibly with a clever substitution. I was especially amazed at how easily it could be solved because integrals with arctan usually give me a lot of trouble.
Also, I think we can extend this to a broader result where we replace $x$ by any arbitrary odd function $g(x)$ and show that 
$$I  = \int\limits_{-a}^{a} \arctan(e^{g(x)})dx = \frac{a \pi}{2}$$
for any odd function $f(x): (-a,a) \to\Bbb R$. Essentially the proof for this would follow the exact same reason as above right? 
So if anyone has another method of computing the original integral I am definitely interested in reading your solutions! Thanks in advance for any input and ideas! Also thanks to Ron Gordon for his nice answer on the question I linked, the answer given there inspired me to look for different ways of trying to solve this integral that I normally would have given up on.
 A: $g(x)$ need to be an odd function, otherwise $f(x)=\arctan(e^{g(x)})-\pi/4$ is not an odd function. 
One simple test is at least $f(x=0)$ should be $0$, which is not satisfied by your choice of $g(x)=1$.
A: I have found another way to integrate this integral, first you have to know this identity, very useful for integrals with arctan
$$\arctan(y)+\arctan(1/y) =\frac{\pi}{2}\tag{1}$$  with $x$ positive or equal to $0$, and
$$\arctan(y)+\arctan(1/y) =\frac{-\pi}{2}$$
for $y$ negative. The first step is separate the integral
$$\int\limits_{-1}^{1} \arctan(e^x)dx=\int\limits_{0}^{1} \arctan(e^x)dx+\int\limits_{-1}^{0} \arctan(e^x)dx\tag{2}$$
after that in the seconnd integral use this substitution ($x=-y$)
and we get
$$\int\limits_{-1}^{0} \arctan(e^x)dx=\int\limits_{0}^{1} \arctan(e^{-y})dy$$
then we can use this and put it in $(1)$ (and replace $x=y$ because is the same what variable we use) and we get
$$\int\limits_{0}^{1} \arctan(e^x)dx+\int\limits_{0}^{1} \arctan(e^{-x})dx$$ after we use $(2)$ and we get
$$\int\limits_{0}^{1} \frac{\pi}{2}dx =\frac{\pi}{2}$$
Then I tryed to generalize and I resolved the same first integral with $x=x^3$, if you want you can try to prove it. Finally we can generalize for $g(x)$ if this function is odd.
$$\int\limits_{-1}^{1} \arctan(e^{g(x)})dx=\int\limits_{0}^{1} \arctan(e^{g(x)})dx+\int\limits_{-1}^{0} \arctan(e^{g(x)})dx$$
we subtitue ($x=-y$) in the second integral and
omitting steps we get
$$\int\limits_{-1}^{1} \arctan(e^{g(x)})dx=\int\limits_{0}^{1} \arctan(e^{g(x)}-e^{g(-x)})dx=\int\limits_{0}^{1} \frac{\pi}{2}dx=\frac{\pi}{2}$$
for $a$ and $-a$ in the limits we can use the same method, if you can't do it I will explain it, thank you for reading and goodbye.
A: $$
\begin{aligned}
I & =\int_{-1}^1 \arctan \left(e^x\right) d x \\
& =\left[x \arctan \left(e^x\right)\right]_{-1}^1-\int_{-1}^1 \frac{x e^x}{1+e^{2 x}} d x \\
& = \arctan e+\arctan (e^{-1})-J
\end{aligned}
$$
Letting $x\mapsto -x$ yields $$
\begin{aligned}
J & =\int_1^{-1} \frac{-x e^{-x}}{1+e^{-2 x}}(-d x) \\
& =-\int_{-1}^1 \frac{x e^x}{e^{2 x}+1} d x \\
& =-J \\
\Rightarrow J & =0 .
\end{aligned}
$$
Hence $$\boxed{I= \arctan e+\arctan (e^{-1})=\frac{\pi}{2}}$$
A: I was working on something similar.
Integration by parts, noticing the initial integral is related to a form of the integral of sech
$I = \int\limits_{-1}^{1} \arctan(e^x)dx$
$= [x \arctan(e^x)] - \int\limits_{-1}^{1} x e^x / (1 + e^{2x})dx$
$= [x \arctan(e^x)] - \int\limits_{-1}^{1} x / (e^{-x} + e^{x})dx$
$= [x \arctan(e^x)] - (1/2)\int\limits_{-1}^{1} x \operatorname{sech}(x)dx$
is then odd I think and equals $[x \arctan(e^x)]$.
I got Nelsen's answer of $\pi/2$, not sure why the $x \arctan$ part works out as that yet, but numerically it did.
A: $$I=\int_{-1}^1\arctan(e^x)dx$$
Substitute $u=e^x$
$$I=\int_{e^{-1}}^e\frac{\arctan{u}}{u}du=\operatorname{Ti}_2(u)\Bigg|_{e^{-1}}^e=\operatorname{Ti}_2(e)-\operatorname{Ti}_2(\frac{1}{e})$$
Derive the following property
$$\arctan{t}+\arctan(1/t)=\frac{\pi}{2}$$
$$\int\frac{\arctan{t}}{t}dt+\int\frac{\arctan{1/t}}{t}dt=\frac{\pi}{2t}$$
$$\operatorname{Ti}_2(t)-\operatorname{Ti}_2(1/t)=\frac{\pi}{2}\ln{t}$$
Plug in t=e into the identity and we get
$$I=\frac{\pi}{2}\ln{e}=\frac{\pi}{2}$$
$\operatorname{Ti}_2(z)$ is the inverse tangent integral
