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$\int_{\alpha+1}^\alpha \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} d x=$
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The correct answer is:
$e^\alpha\left(\frac{e-2}{2}\right)$
$\begin{aligned} I & =\int_{\alpha+1}^\alpha \frac{e^x(\alpha-x)}{(x-\alpha+1)^2} d x=\int_{\alpha+1}^\alpha \frac{e^x(\alpha-x-1+1)}{(x-\alpha+1)^2} d x \\ & =\int_{\alpha+1}^\alpha e^x\left[\frac{1}{(x-\alpha+1)^2}-\frac{1}{(x-\alpha+1)}\right] d x \\ & =\left[-\frac{e^x}{(x-\alpha+1)}\right]_{\alpha+1}^\alpha \\ & =-e^\alpha+\frac{e^{\alpha+1}}{2}=e^\alpha\left[\frac{e-2}{2}\right]\end{aligned}$
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