$\arctan{x}+\arctan{y}$ from integration

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*}