I solved this integral a couple of years ago and I had this solution typed out in $\LaTeX$ already. The solution is not conventional, so I think it's worth sharing!
First, substitute the series $\log(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}x^n}{n}.$
This series is uniformly convergent on $[0,1]$, so we can interchange the sum and the integral. We get
$I=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 \frac{x^n}{1+x^2}\: \mbox{d}x. = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n}C_n$
where $C_n=\int_0^1 \frac{x^n}{1+x^2} \mbox{d}x.$
Now since $x^{n-2}-\frac{x^{n-2}}{1+x^2}=\frac{x^n}{1+x^2},$
we have, integrating this equation on $[0,1]$, $\frac{1}{n-1}-C_{n-2}=C_n.$ Hence we have a recurrence relation for the $C_n$'s. Let's see what this gives. We have $C_0=\int_0^1 \frac{ \mbox{d}x}{1+x^2} = \arctan(1) = \frac{\pi}{4},$ and
$C_1=\int_0^1 \frac{x\ \mbox{d}x}{1+x^2} = \frac{1}{2}\log2.$
Now using the recurrence we find
$C_0=\frac{\pi}{4}$
$C_1=\frac{1}{2}\log2$
$C_2=1-\frac{\pi}{4}$
$C_3=\frac{1}{2}-\frac{1}{2}\log2$
$C_4 = -1+\frac{1}{3} +\frac{\pi}{4}$
$C_5=-\frac{1}{2}+\frac {1}{4} +\frac{1}{2}\log 2$
$C_6=1-\frac{1}{3}+\frac{1}{5} - \frac{\pi}{4}$
...
Now if we define
$A_n = \sum_{k=1}^n\frac{(-1)^{k-1}}{2k},$ $B_n = \sum_{k=1}^n\frac{(-1)^{k-1}}{2k-1},$ it is easy to see by induction that $C_{2n} = (-1)^n\left(\frac{\pi}{4}-B_n\right)$ and
$C_{2n-1} = (-1)^{n-1}\left(\frac{\log 2}{2} -A_{n-1}\right).$
Notice that $A_n \rightarrow \frac{1}{2}\log2$ as $n\rightarrow \infty$ [recall the series expansion of $\log(1+x)$], and that $B_n \rightarrow \frac{\pi}{4}$ as $n\to \infty$ [recall the series expansion of $\arctan x$, which you can get by integrating $(1+x^2)^{-1}$].
Let us examine the partial sums
$I_{2N}=\sum_{n=1}^{2N} \frac{(-1)^{n-1}}{n}C_n.$
We can split this into the odd-labeled terms and the even-labeled terms, as
$I_{2N}= \sum_{n=1}^{N} \frac{C_{2n-1}}{2n-1} - \sum_{n=1}^{N} \frac{C_{2n}}{2n}.$
Now we can substitute the values of $C_{2n}$ and $C_{2n-1}$ obtained before. First, let us look at the sum of the even-labeled terms. We have
$\sum_{n=1}^{N} \frac{C_{2n}}{2n} =\sum_{n=1}^{N} \frac{ (-1)^{n}}{2n}\left(\frac{\pi}{4}-B_n\right)=-\frac{\pi}{4}A_N +\sum_{n=1}^N\frac{(-1)^{n-1}}{2n}B_n $
Now let us recall Cauchy's partial summation formula. For sequences $\{a_n\}, \{b_n\}$, we denote $\{\Delta a_n\}$ the sequence of forward differences $\Delta a_n = a_{n+1}-a_n$. Then we have
$\sum_{n=1}^N b_n \Delta a_{n-1} = b_Na_N - b_1a_0 -\sum_{n=1}^{N-1} \Delta b_n a_n.$
Remark that $\Delta A_{n-1} = \frac{(-1)^{n-1}}{2n}$. Also, $A_0=0$. Hence,
$-\frac{\pi}{4}A_N +\sum_{n=1}^N\Delta A_{n-1}B_n = -\frac{\pi}{4}A_N +B_NA_N -\sum_{n=1}^{N-1}\Delta B_n A_n.$
(Call this sum (1)).
Now we go back to the sum of the odd-labeled terms. We have, in a similar fashion,
$\sum_{n=1}^{N} \frac{C_{2n-1}}{2n-1} = \sum_{n=1}^{N} \frac{(-1)^{n-1}}{2n-1}\left(\frac{\log 2}{2} - A_{n-1}\right) = \frac{\log 2}{2} B_N - \sum_{n=1}^N \Delta B_{n-1}A_{n-1}$ $= \frac{\log 2}{2} B_N - \sum_{n=0}^{N-1} \Delta B_{n}A_{n}.$ (Call this sum (2)).
Now, subtracting (1) from (2), we get
$I_{2N} = \frac{\log 2}{2}B_N + \frac{\pi}{4}A_N - B_NA_N +\Delta B_0 A_0$ $= \frac{\log 2}{2}B_N + \frac{\pi}{4}A_N - B_NA_N$
Now as $N\to \infty$, since $A_N \to \frac{1}{2}\log 2$ and $B_N \to \frac{\pi}{4}$, we have
$I = \int_0^1\frac{\log(1+x)}{1+x^2}\: \mbox{d}x = \lim_{N\to \infty} I_{2N} = \frac{\pi \log 2}{8}.$