<s>[INST] Show that $\int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2} = \frac{\pi^2}{8}$ $\displaystyle \int_0^1 \tan^{-1}\left(\frac{3(1+x)}{1-2x-x^2}\right) \frac{dx}{1+x^2} = \frac{\pi^2}{8}$
I had been told that $x =\displaystyle \frac{1-y}{1+y}$ solves it, but it takes me back to where I started.
 [/INST] In fact, the correct answer is 0. As @JimmyK4542 computed, we know from the substitution $x \mapsto (1-x)/(1+x)$ that
$$ I = \int_{0}^{1} \arctan\left( \frac{3(1+x)}{1-2x-x^2} \right) \, \frac{dx}{1+x^2} = \int_{0}^{1} \arctan\left( - \frac{3(1+x)}{1-2x-x^2} \right) \, \frac{dx}{1+x^2}. $$
But the identity $\arctan(-x) = -\arctan x$ is true for any $x \in \Bbb{R}$. Consequently
$$ I = -I $$
and hence $I = 0$. So why we have vanishing quantity here? This is because the function satisfies
$$ x \mapsto \arctan\left(\frac{3(1+x)}{1-2x-x^2}\right) \quad \text{is} \quad \begin{cases}
\text{positive when } 0 < x < \sqrt{2}-1 \\
\text{negative when } \sqrt{2}-1 < x < 1.
\end{cases} $$
So the graph of this function looks like this:

Somehow magically, the contribution from the positive part and that from the negative part cancel out each other. This observation is also supported by numerical calculation:

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