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If $\mathrm{b}=\int_{0}^{1} \frac{\mathrm{e}^{t}}{\mathrm{t}+1} d t$, then $\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{-1}}{\mathrm{t}-\mathrm{a}-1}$ is
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The correct answer is:
$-\mathrm{be}^{-\mathrm{a}}$
$\int_{a-1}^{a} \frac{e^{-t}}{t-a-1} d t$
$=\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{-(2 \mathrm{a}-1-\mathrm{t})}}{2 \mathrm{a}-1-\mathrm{t}-\mathrm{a}-1} \mathrm{dt}$
$=\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{\mathrm{t}-2 \mathrm{a}+1}}{\mathrm{a}-2-\mathrm{t}} \mathrm{dt}$
$t-(a-1)=x \quad \Rightarrow d t=d x$
$=\int_{0}^{1} \frac{e^{x} \cdot e^{-a}}{-x-1} d x$
$=-e^{-a} \int_{0}^{1} \frac{e^{x}}{x+1} d x=-b e^{-a}$
$=\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{-(2 \mathrm{a}-1-\mathrm{t})}}{2 \mathrm{a}-1-\mathrm{t}-\mathrm{a}-1} \mathrm{dt}$
$=\int_{\mathrm{a}-1}^{\mathrm{a}} \frac{\mathrm{e}^{\mathrm{t}-2 \mathrm{a}+1}}{\mathrm{a}-2-\mathrm{t}} \mathrm{dt}$
$t-(a-1)=x \quad \Rightarrow d t=d x$
$=\int_{0}^{1} \frac{e^{x} \cdot e^{-a}}{-x-1} d x$
$=-e^{-a} \int_{0}^{1} \frac{e^{x}}{x+1} d x=-b e^{-a}$
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