Seeking methods to solve: $\int_{0}^{1} \frac{1}{1 + \arctan(x)} \:dx$

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.