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.
 A: I have an answer, though it is without using differentiation under integration
$$I=\int_{0}^1 \frac{\tan^{-1}x}{1+x}dx=\int_{0}^{\pi/4}\frac{\theta \sec^2\theta}{1+\tan \theta}d\theta\\ =\int_{0}^{\pi/4}\frac{(\pi/4-\theta)\sec^2(\pi/4-\theta)}{1+\frac{1-\tan\theta}{1+\tan\theta}}d\theta\\=\int_{0}^{\pi/4}\frac{(\pi/4-\theta)}{1+\tan\theta}\sec^2\theta d\theta\\ \Rightarrow 2I=\pi/4\int_{0}^{\pi/4}\frac{\sec^2\theta }{1+\tan\theta}d\theta=\pi/4\ln 2\\ \Rightarrow I=\pi/8\ln 2$$
A: 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
$$
\frac{1}{1+x} \frac{p}{1+p^2 x^2}=\frac{p}{\left(p^2+1\right) (x+1)}+\frac{p^3-p^3 x}{\left(p^2+1\right) \left(p^2 x^2+1\right)}
$$
A: Another answer without using differentiation integration using a change of variable of type homograph:
$$x=\frac{1-t}{1+t}, dx=\frac{-2}{(1+t)^2},0\leq t\leq1.$$
$$I=\int_{0}^1 \frac{\arctan x}{1+x}dx= \int_{1}^0 \frac{\arctan \frac{1-t}{1+t} }{1+\frac{1-t}{1+t}}\frac{-2}{(1+t)^2}dx=\int_{0}^1 \frac{\arctan \frac{1-t}{1+t} }{1+t}dx=$$
$$=\int_{0}^1 \frac{\frac{\pi}{4}-\arctan t}{1+t}dx.$$
$$2I=\frac{\pi}{4}\int_{0}^1 \frac{1}{1+t}dx, I=\frac{\pi}{8}\ln 2.$$
A: $$I(a) = \int \frac{\arctan (ax)}{x+1}dx \Rightarrow I'(a) = \int \frac{x}{(x + 1)(a^2x^2+1)} dx$$
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 $\int\frac{1}{a^2x^2+1}dx = \frac{\arctan (ax)}{a}$. 
A: Integrating by parts:
$$\begin{eqnarray*} I = \int_{0}^{1}\frac{\arctan x}{x+1}\,dx &=& \left.\log(x+1)\arctan(x)\right|_{0}^{1}-\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,dx\\&=&\frac{\pi}{4}\log 2-\int_{0}^{\pi/4}\log(1+\tan\theta)\,d\theta \tag{1}\end{eqnarray*}$$
but since $\tan\left(\frac{\pi}{4}-\theta\right)=\frac{1-\tan\theta}{1+\tan\theta}$ we have:
$$ I = \frac{\pi}{4}\log 2 - \int_{0}^{\pi/4}\log 2\,d\theta + \int_{0}^{\pi/4}\log(1+\tan\theta)\,d\theta = \int_{0}^{\pi/4}\log(1+\tan\theta)\,d\theta\tag{2}$$
and by summing $(1)$ and $(2)$ we get $2I=\frac{\pi}{4}\log 2$, so:
$$ I = \color{red}{\frac{\pi}{8}\log 2}.\tag{3}$$
A: If we consider $$I(a) = \int_{0}^{1}\frac{\arctan (ax)}{x+1}\mathrm{d}x$$ and then we apply the Differentiation under integral sign Method finding $I'(a)$, it may be very difficult to integrate back and find $I(a)$.
So I suggest to proceed in a bit different way.
Integrating by parts, we have
$$I = \int_{0}^{1}\frac{\arctan x}{x+1}\mathrm{d}x = \left.\log(x+1)\arctan(x)\right|_{0}^{1}-\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\mathrm{d}x=\frac{\pi}{4}\log 2-J$$
where 
$$
J=\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\mathrm{d}x.
$$
Consider the parametric integral
$$
J(a)=\int_{0}^{a}\frac{\log(1+ax)}{1+x^2}\mathrm{d}x
$$
so that the original integral is
$$
I=\frac{\pi}{4}\log 2-J(1).
$$
Differentiating we have
$$
\begin{align}
J'(a)&=\int_{0}^{a}\frac{x}{(1+x^2)(1+ax)}\mathrm{d}x+\frac{\log(1+a^2)}{1+a^2}\\
&=\int_{0}^{a}\left[\frac{-a}{1+a^2}\frac{1}{1+ax}+\frac{1}{1+a^2}\frac{x+a}{1+x^2}\right]\mathrm{d}x+\frac{\log(1+a^2)}{1+a^2}\\
&=\frac{-1}{1+a^2}\log(1+a^2)+\frac{1}{1+a^2}\frac{1}{2}\log(1+a^2)+\frac{a}{1+a^2}\arctan a+\frac{\log(1+a^2)}{1+a^2}\\
&=\frac{1}{1+a^2}\frac{1}{2}\log(1+a^2)+\frac{a}{1+a^2}\arctan a
\end{align}
$$
so that
$$
J(a)=\frac{1}{2}\int \frac{1}{1+a^2}\log(1+a^2)\,\mathrm{d}a+\int \frac{a}{1+a^2}\arctan a \,\mathrm{d}a
$$
and integrating by parts the second integral we find
$$
J(a)=\frac{1}{2}\int \frac{\log(1+a^2)}{1+a^2}\,\mathrm{d}a+(\arctan a)\frac{1}{2}\log(1+a^2)-\frac{1}{2}\int \frac{\log(1+a^2)}{1+a^2}\,\mathrm{d}a+C
$$
that is
$$
J(a)=(\arctan a)\frac{1}{2}\log(1+a^2)+C
$$
where $C$ is a constant; observing that $J(0)=0$, we find $C=0$.
So we have
$$
J(a)=(\arctan a)\frac{1}{2}\log(1+a^2)
$$
and then
$$
I=\frac{\pi}{4}\log 2-J(1)=\frac{\pi}{4}\log 2-\arctan(1)\frac{1}{2}\log(2)=\frac{\pi}{4}\log 2-\frac{\pi}{8}\log 2
$$
that is
$$
I=\frac{\pi}{8}\log 2.
$$
A: Let
$$I=\int_0^1\frac{\arctan(ax)}{1+x}dx,\quad \quad I(a)=\int_0^1\frac{\arctan(ax)}{1+x}dx$$
and note that $I(0)=0$ and $I(1)=I.$
$$I=\int_0^1 I'(a)da=\int_0^1\left(\int_0^1\frac{x}{(1+x)(1+a^2 x^2)}dx\right)da$$
$$=\int_0^1\frac{\arctan(a)}{a}da-\underbrace{\int_0^1\frac{a\,\arctan(a)}{1+a^2}da}_{IBP}+\frac12\int_0^1\frac{\ln(1+a^2)}{1+a^2}da-\int_0^1\frac{\ln(2)}{1+a^2}da$$
$$=G-\frac{\pi}{8}\ln(2)+\frac12\int_0^1\frac{\ln(1+a^2)}{1+a^2}da+\frac12\int_0^1\frac{\ln(1+a^2)}{1+a^2}da-\frac{\pi}{4}\ln(2)$$
$$=G-\frac{3\pi}{8}\ln(2)+\int_0^1\frac{\ln(1+a^2)}{1+a^2}da$$
$$=G-\frac{3\pi}{8}\ln(2)+\frac{\pi}{2}\ln(2)-G$$
$$=\frac{\pi}{8}\ln(2)$$

$$J=\int_0^1\frac{\ln(1+x^2)}{1+x^2}dx=\underbrace{\int_0^\infty\frac{\ln(1+x^2)}{1+x^2}dx}_{x=\tan \theta}-\underbrace{\int_1^\infty\frac{\ln(1+x^2)}{1+x^2}dx}_{x=1/y}$$
$$=-2\underbrace{\int_0^{\pi/2}\ln(\cos x)dx}_{\text{Fourier series}}-\int_0^1\frac{\ln(\frac{1+x^2}{x^2})}{1+x^2}dx$$
$$=-2(-\frac{\pi}{2}\ln(2))-J-2G$$
$$\Longrightarrow J=\frac{\pi}{2}\ln(2)-G$$
