I attempted to evaluate the following integral, $$\int_{0}^{1} \frac{\ln (1+x)}{1+x^2}\, dx$$ although I know it can be evaluated using Feynman's Technique of Integration, Trigonometric Substitution, etc. I used the following method, but I was stuck at computing the double summation: $$\begin{align*} \int_{0}^{1} \frac{\ln (1+x)}{1+x^2} \, dx &= \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^n x^{2n} \ln (1+x) \, dx \\ &= \sum_{n=0}^{\infty} (-1)^n\int_{0}^{1} x^{2n}\ln (1+x)\, dx \\ &= \sum_{n=1}^{\infty} (-1)^n \left(\frac{\ln (1+x)x^{2n+1}}{2n+1}\Bigg\rvert_{0}^{1}- \frac{1}{2n+1}\int_{0}^{1}\frac{x^{2n+1}}{1+x}\, dx\right) \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n \ln 2}{2n+1} -\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1}\left(x^{2n}-x^{2n-1}+\cdots +x^2-x+1-\frac{1}{1+x}\right)\, dx \\&= \frac{\pi}{2} \ln 2 -\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \left(\frac{1}{2n+1}-\frac{1}{2n}+\cdots +\frac{1}{3}-\frac{1}{2}+1\right) \\ &=\frac{\pi}{2} \ln 2 -\sum_{n=0}^{\infty} \sum_{m=0}^{2n} \frac{(-1)^n}{2n+1}\frac{(-1)^m}{m+1}. \end{align*}$$ I tried to apply the fact that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots +\frac{1}{2n}$ which makes the double summation into $$\sum_{n=0}^{\infty}\sum_{m=n}^{2n} \frac{(-1)^n}{2n+1}\frac{1}{m+1}=\sum_{n=0}^{\infty} \sum_{m=0}^{n}\frac{(-1)^n}{2n+1}\frac{1}{m+n+1}$$ and switching order of summation gives $$\sum_{m=0}^{\infty} \sum_{n=m}^{\infty} \frac{(-1)^n}{(2n+1)(m+n+1)}=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty} \frac{(-1)^{n+m}}{(2(n+m)+1)(2m+n+1)}$$ but this doesn't make it simpler, and I am not sure if there's a way of calculating the sum. I would appreciate if anyone has a way to evaluate this.
Note: The integral $\displaystyle \int_{0}^{1}\frac{\ln(1+x)}{1+x^2}\, dx$ has the value, $\frac{\pi}{8}\ln 2$, so this means we should have $$\sum_{n=0}^{\infty} \sum_{m=0}^{2n} \frac{(-1)^n}{2n+1}\frac{(-1)^m}{m+1}=\frac{3}{8}\ln 2.$$
