A) \[\frac{e}{4}\]
B) \[\frac{e}{4}-1\]
C) \[\frac{e}{4}+1\]
D) None of these
Correct Answer: B
Solution :
\[\int_{0}^{1}{\frac{{{e}^{x}}(x-1)}{{{(x+1)}^{3}}}}dx=\int_{0}^{1}{\frac{{{e}^{x}}(x+1-2)}{{{(x+1)}^{3}}}\,}dx\] \[\int_{0}^{1}{\frac{{{e}^{x}}}{{{(x+1)}^{2}}}}dx-2\int_{0}^{1}{\frac{{{e}^{x}}}{{{(x+1)}^{3}}}}dx=\left[ \frac{{{e}^{x}}}{{{(x+1)}^{2}}} \right]_{0}^{1}=\frac{e}{4}-1\].
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