$\arctan{x}+\arctan{y}$ 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...
 A: 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*}
A: Another change of variables that works, very similar to that in the answer by @DonaldSplutterwit, is the following:
$$
t=f(u)=\frac{1-u\sigma}{u+\sigma}\,,\qquad\text{with}\ \ \sigma(x,y)=\frac{1-xy}{x+y}\,.
$$
It is more symmetric, since it works both for 
$$
\int_x^{1/\sigma}\frac{dt}{1+t^2}=\int_0^y\frac{du}{1+u^2}
$$
and for
$$
\int_y^{1/\sigma}\frac{dt}{1+t^2}=\int_0^x\frac{du}{1+u^2}\,.
$$
Indeed,
$$
f(0)=\frac{1}{\sigma}\,,\qquad
f(x)=y\,,\qquad f(y)=x
$$
and
$$
dt=-\frac{1+\sigma^2}{(u+\sigma)^2}du\,,\qquad 
\frac{1}{1+t^2}=\frac{(u+\sigma)^2}{(1+\sigma^2)(1+u^2)}\,.
$$
It also has the property of reducing to the inversion as $xy\to1^-$, namely $\sigma\to0^+$, since
$$
f(u)\big|_{\sigma=0}=\frac{1}{u}\,,
$$
and we get back
$$
\int_{x}^\infty \frac{dt}{1+t^2} = \int_{0}^{\frac{1}{x}}\frac{du}{1+u^2}\,.
$$
In fact, $f$ is also an involution $f(f(u))=u$ and also allows to run the proof "forward" in the following way
$$
\int_0^x\frac{dt}{1+t^2}+\int_0^y\frac{dt}{1+t^2}=\int_0^x\frac{dt}{1+t^2}+\int_x^{\frac{1}{\sigma}}\frac{du}{1+u^2}=\int_0^{\frac{1}{\sigma}}\frac{dt}{1+t^2}\,,
$$
where we let $t=f(u)$ in the second integral.
A: hint
If we put
$$t=\frac{x+y}{1-xy}u$$
the left integral becomes
$$\int_0^1\frac{1}{1+(\frac{x+y}{1-xy})^2u^2}\frac{x+y}{1-xy}du$$
A: Fixing $y$, define $f(x):=\arctan x+\arctan y-\arctan\frac{x+y}{1-xy}$ so $f(0)=0$ and$$\begin{align}f^\prime(x)&=\frac{1}{1+x^2}-\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\partial_x\frac{x+y}{1-xy}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1-xy-(x+y)(-y)}{(1-xy)^2}\\&=\frac{1}{1+x^2}-\frac{(1-xy)^2}{(1+x^2)(1+y^2)}\frac{1+y^2}{(1-xy)^2}\\&=0,\end{align}$$i.e. $f(x)=0$ for all $x$.
