$\arctan{x}+\arctan{y}$ from integration I was trying to derive the property 
$$\arctan{x}+\arctan{y}=\arctan{\frac{x+y}{1-xy}}$$ 
for $x,y>0$ and $xy<1$ from the integral representation
$$
\arctan{x}=\int_0^x\frac{dt}{1+t^2}\,.
$$
I am aware of "more trigonometric" proofs, for instance using that $\tan{(\alpha+\beta)}=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$, but I was willing to see if there is a proof that uses more directly the properties of the integral representation. For instance, if $x>0$, one immediately gets
$$\begin{aligned}
\arctan{x}+\arctan\frac{1}{x}
&=\int_0^x\frac{dt}{1+t^2} + \int_0^{\frac{1}{x}}\frac{dt}{1+t^2}\\
&=
\int_0^x\frac{dt}{1+t^2}+\int_x^\infty\frac{dt}{1+t^2}\\
&=\int_0^\infty \frac{dt}{1+t^2} = \frac{\pi}{2}
\end{aligned}$$
sending $t\to\frac{1}{t}$ in the second integral.
Similarly I tried considering
$$
\int_0^x\frac{dt}{1+t^2} + \int_0^y\frac{dt}{1+t^2}=(x+y)\int_0^1\frac{1+xyt^2}{1+(x^2+y^2)t^2+x^2y^2t^4}\ dt
$$
after rescaling $t\to xt$ and $t\to yt$. On the other hand, via a similar rescaling $t\to \frac{x+y}{1-xy}t$, we have
$$
\int_0^\frac{x+y}{1-xy}\frac{dt}{1+t^2}
=
(x+y)\int_0^1\frac{1-xy}{(1-xy)^2+(x+y)^2t^2}\ dt\,.
$$
By a clever choice of variable it should (must?) be possible to see that these integrals are actually the same, but I can't figure it out...

We want show that
\begin{eqnarray*}
\int_x^{ \frac{x+y}{1-xy}} \frac{dt}{1+t^2} = \int_{0}^{y} \frac{du}{1+u^2}
\end{eqnarray*}
that's to say the LHS is actually independent of $x$.
The substitution 
\begin{eqnarray*}
t=x+ \frac{u(1+x^2)}{1-ux}
\end{eqnarray*}
will do the trick.
The limits are easily checked and we have 
\begin{eqnarray*}
dt= \frac{1+x^2}{(1-ux)^2} du.
\end{eqnarray*}
The rest is a little bit of algebra.
Note the similarity with $ \ln(a)+\ln(b) = \ln(ab)$
\begin{eqnarray*}
\int_{1}^{a} \frac{dt}{t} +\int_{1}^{b} \frac{dt}{t} = \int_{1}^{ab} \frac{dt}{t}.
\end{eqnarray*}
And $ u=at $
\begin{eqnarray*}
\int_{1}^{b} \frac{dt}{t} = \int_{a}^{ab} \frac{du}{u}.
\end{eqnarray*}
