Question

Va rog sa ma ajutati la ex din poza.

Answer 1

$\displaystyle \int_{1}^2\dfrac{1}{|x-a|+1}\, dx=\ln a -\ln(a-1)\\\\\\\Rightarrow \int \dfrac{1}{|x-a|+1}\, dx =-\ln(a+1-x)+C\,\,\Bigg|\dfrac{d}{dx} \\ \\ \\ \Rightarrow \dfrac{1}{|x-a|+1} =-\dfrac{-1}{a+1-x} \\ \\\\ \Rightarrow \dfrac{1}{|x-a|+1}= \dfrac{1}{-(x-a)+1}\\ \\ \\ \Rightarrow x-a \leq 0 \Rightarrow [1,2] - a \leq 0 \Rightarrow [1,2]\leq a\\\\ \\ \Rightarrow \boxed{a\geq 2}$