If \[A=\int\limits_{0}^{1}{\frac{{{e}^{t}}}{t+1}dt,}\] then \[\int\limits_{0}^{1}{{{e}^{t}}\log (1+t)dt}\] in terms of A equals

If \[A=\int\limits_{0}^{1}{\frac{{{e}^{t}}}{t+1}dt,}\] then \[\int\lim

1.	If \[A=\int\limits_{0}^{1}{\frac{{{e}^{t}}}{t+1}dt,}\] then \[\int\limits_{0}^{1}{{{e}^{t}}\log (1+t)dt}\] in terms of A equals
A.	\[e\log (A)\]
B.	\[\frac{e}{2}-A\]
C.	\[e-l-\frac{A}{2}\]
D.	\[\frac{e}{2}-l-A\]
Answer» E.

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If \[A=\int\limits_{0}^{1}{\frac{{{e}^{t}}}{t+1}dt,}\] then \[\int\limits_{0}^{1}{{{e}^{t}}\log (1+t)dt}\] in terms of A equals

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