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.

I have an answer, though it is without using differentiation under integration
$$I=\int_{0}^1 \frac{\tan^{-1}x}{1+x}dx=\int_{0}^{\pi/4}\frac{\theta \sec^2\theta}{1+\tan \theta}d\theta\\ =\int_{0}^{\pi/4}\frac{(\pi/4-\theta)\sec^2(\pi/4-\theta)}{1+\frac{1-\tan\theta}{1+\tan\theta}}d\theta\\=\int_{0}^{\pi/4}\frac{(\pi/4-\theta)}{1+\tan\theta}\sec^2\theta d\theta\\ \Rightarrow 2I=\pi/4\int_{0}^{\pi/4}\frac{\sec^2\theta }{1+\tan\theta}d\theta=\pi/4\ln 2\\ \Rightarrow I=\pi/8\ln 2$$
