Integration by differentiating under the integral sign $I = \int_0^1 \frac{\arctan x}{x+1} dx$ $$I = \int_0^1 \frac{\arctan x}{x+1} dx$$
I spend a lot of my time trying to solve this integral by differentiating under the integral sign, but I couldn't get something useful. I already tried:
$$I(t) = \int_0^1 e^{-tx}\frac{\arctan x}{x+1} dx ; \int_0^1 \frac{(t(x+1)-x)\arctan x}{x+1} dx ; \int_0^1 \frac{\arctan tx}{x+1} dx ; \int_0^1 \frac{\ln(te^{\arctan x})}{x+1} dx $$
And something similar. A problem is that we need to get +constant at the very end, but to calculate that causes same integration problems.
To these integrals:
$$I(t) = \int_0^1 e^{-tx}\frac{\arctan x}{x+1} dx ; \int_0^1 \frac{\arctan tx}{x+1} dx$$
finding constant isn't problem, but to solve these integrals by itself using differentiating under the integral sign is still complexive.
Any ideas? I know how to solve this in other ways (at least one), but I particularly interesting in differentiating.

Another answer without using differentiation integration using a change of variable of type homograph:
$$x=\frac{1-t}{1+t}, dx=\frac{-2}{(1+t)^2},0\leq t\leq1.$$
$$I=\int_{0}^1 \frac{\arctan x}{1+x}dx= \int_{1}^0 \frac{\arctan \frac{1-t}{1+t} }{1+\frac{1-t}{1+t}}\frac{-2}{(1+t)^2}dx=\int_{0}^1 \frac{\arctan \frac{1-t}{1+t} }{1+t}dx=$$
$$=\int_{0}^1 \frac{\frac{\pi}{4}-\arctan t}{1+t}dx.$$
$$2I=\frac{\pi}{4}\int_{0}^1 \frac{1}{1+t}dx, I=\frac{\pi}{8}\ln 2.$$
