<br /> \begin{cases}<br /> f(1)=0\\<br /> f(e^2)=0<br /> \end{cases}<br /> \implies\begin{cases}<br /> 1\ln 1-\frac{1}{2}\cdot 1+B=0\\<br /> A\ln e^2-\frac{1}{2}e^2+B=0<br /> \end{cases}<br /> \implies<br /> \begin{cases}<br /> B=\displaystyle\frac{1}{2}\\<br /> A=\displaystyle\frac{e^2-1}{4{\color{red}\ln e}}<br /> \end{cases}<br />
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