<s>[INST] Seeking methods to solve: $\int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$ I've been playing with the following definite integral and was wondering if anyone knew of any methods to solve?
$$I = \int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$$
 [/INST] The approach I took:
First let $u = \tan(x)$ to yield:
$$I = \int_{0}^{\frac{\pi}{4}} \frac{\tan^2(u) + 1}{1 + u} \:du$$
Unfortunately I had no luck with my usual tactics, so I decided to use the Taylor series of $\tan^2(u)$ at $u = 0$ which I was greatly helped with here. 
We find that:
$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} $$
Which holds for $|u| < \frac{\pi}{2}$. As the domain of the integral is within that, we can use this expansion (I believe). 
Hence, we arrive at, 
$$I = \int_{0}^{\frac{\pi}{4}} \frac{\sum_{n = 1}^{\infty} C_n u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{1 + u} \:du = \sum_{n = 1}^{\infty}C_n\: F_n $$
Where 
$$ C_n = \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} ,\qquad F_n = \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du$$
Taking the Taylor expansion for $\frac{1}{1 + u}$ at $u = 0$ (as given here) we can evaluate $F_n$:
\begin{align}
F_n &= \int_{0}^{\frac{\pi}{4}} \frac{u^{2n - 2}}{u + 1} \: du \\
&= \left[\ln|u + 1| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k u^k \right]_{0}^{\frac{\pi}{4}} \\
&= \ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk}
\end{align}
From which we arrive at:
\begin{align}
I =  \sum_{n = 1}^{\infty}  C_n\:F_n &=  \sum_{n = 1}^{\infty}  C_n \left[\ln\left|\frac{\pi}{4} + 1 \right| + \sum_{k = 1}^{2n - 2} \left(-1\right)^k \frac{\pi^k}{4^kk} \right] \\
&= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty}  C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk}
\end{align}
Recall that
$$ \tan^{2}(u) + 1 = \sum^{\infty}_{n=1} \frac{B_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)}{(2n)!} u^{2n-2} = \sum_{n = 1}^{\infty} C_n u^{2n - 2} $$
Hence, 
$$ \tan^{2}(1) + 1 = \sec^{2}(1) = \sum_{n = 1}^{\infty} C_n$$
Thus, 
\begin{align}
 I &= \ln\left|\frac{\pi}{4} + 1 \right| \sum_{n = 1}^{\infty}  C_n + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n - 2} C_n \left(-1\right)^k \frac{\pi^k}{4^kk} \\
&= \ln\left|\frac{\pi}{4} + 1 \right|\sec^2(1) + \sum_{n = 1}^{\infty}\sum_{k = 1}^{2n -2}\frac{\left(-1\right)^kB_{2n} (-4)^n \left(1-4^n\right)\left(2n - 1\right)\pi^k}{(2n)!4^kk} 
\end{align}
However at this stage, I'm lost as to how this could be simplified. 
</s>