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.

When we want to integrate
$$
\int_a^b f(x) \arctan x dx
$$
where $f$ is a relatively simple function (e.g. a rational function), we let
$$
J(p) := \int_a^b f(x) \arctan px dx
$$
so
$$
J'(p) = \int_a^b f(x) \frac{p}{1+p^2 x^2} dx.
$$
The reason why this works is that we know $J(0) = 0$, so
$$
J(1) = \int_0^1 J'(p) dp.
$$
This should take care of your problem with the '+constant'.
To calculate 
$$
J'(p) = \int_a^b f(x) \frac{p}{1+p^2 x^2} dx
$$
in your case,  it may help to know that
$$
\frac{1}{1+x} \frac{p}{1+p^2 x^2}=\frac{p}{\left(p^2+1\right) (x+1)}+\frac{p^3-p^3 x}{\left(p^2+1\right) \left(p^2 x^2+1\right)}
$$
