Show that \(\displaystyle\int_0^1 \frac{dx}{(ax + b(1-x))^2} = \frac{1}{ab}\), \(a, b \in \mathbb{R}\).
Erratum: as stated, with arbitrary \(a, b \in \mathbb{R}\), the identity is false. For \(a = 1\), \(b = -1\) the integrand is \(\frac{1}{(2x-1)^2}\), which is not integrable near \(x = \frac{1}{2}\), so the integral diverges, while \(\frac{1}{ab} = -1\). More generally, if \(ab < 0\) then \(ax + b(1-x)\) vanishes at \(x = \frac{b}{b-a} \in (0,1)\) and the integral diverges, and if \(ab = 0\) the right-hand side is undefined. The identity holds under the corrected hypothesis \(ab > 0\), that is, \(a\) and \(b\) nonzero of the same sign.