<s>[INST] 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.
 [/INST] $$I(a) = \int \frac{\arctan (ax)}{x+1}dx \Rightarrow I'(a) = \int \frac{x}{(x + 1)(a^2x^2+1)} dx$$
Now $$\frac{x}{(x + 1)(a^2x^2+1)}  =\frac{A}{(x + 1)} + \frac{Bx + C}{(a^2x^2+1)}$$
then $A = \frac{-1}{a^2 + 1}$, $B = \frac{a^2}{a^2 + 1}$ and $C = \frac{1}{a^2+1}$. From this we have 
$$\begin{align}I'(a) &= \int \frac{-1}{(a^2 + 1)(x+1)}dx + \int\frac{a^2x + 1}{(a^2 + 1)^2(a^2x^2+1)}dx \\&=\frac{-1}{(a^2+1)}\int \frac{1}{(x+1)}dx + \frac{1}{(a^2+1)^2}\Bigg[\int \frac{a^2x}{a^2x^2+1}dx +\int\frac{1}{a^2x^2+1}dx\Bigg]\end{align}$$
Where $\int \frac{a^2x}{a^2x^2+1}dx = x - \frac{\ln (a^2x^2 + 1)}{a}$ and $\int\frac{1}{a^2x^2+1}dx = \frac{\arctan (ax)}{a}$. 
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