$$I=\int_0^1\frac{\ln(x+1)}{x^2+1}dx$$ $u=x+1$ so $du=dx$ $$I=\int_1^2\frac{\ln(u)}{(u-1)^2+1}du$$ $v=\ln(u)$ so $du=udv$ $$I=\int_0^{\ln(2)}\frac{ve^v}{(e^v-1)^2+1}dv$$ $e^v-1=\tan(w)$ so $dv=\frac{\sec^2(w)dw}{e^v}$ $$I=\int_0^{\pi/4}vdw=\int_0^{\pi/4}\ln(\tan(w)+1)dw$$ However this does not seem to make the problem any easier, I was thinking about using Feynman's trick, letting: $$I(a)=\int_0^1\frac{\ln(x+1)}{x^a+1}dx$$ but this also seems to complicate things more.
I also thought of using a different new variable: $$I(b)=\int_0^1\frac{\ln(x+1)}{(x+1)^a}dx$$ $$I'(b)=\int_0^1-\frac{\ln^2(x+1)}{(x+1)^a}dx$$ but this had the opposite effect to what I'm looking for (eliminate the $\ln(x+1)$)
EDIT
The hint given is: $$\tan(x\pm y)=\frac{\tan(x)\pm\tan(y)}{1-\mp\tan(x)\tan(y)}$$ and I see the potential link to $\arctan$ but not $\tan$
