If \[\int{\frac{x{{e}^{x}}}{\sqrt{1+{{e}^{x}}}}dx=f(x)\sqrt{1+{{e}^{x}}}-2\log \,\,g(x)+C,}\] then

\[g(x)=\frac{\sqrt{1+{{e}^{x}}}+1}{\sqrt{1+{{e}^{x}}}-1}\]

\[g(x)=\frac{\sqrt{1+{{e}^{x}}}-1}{\sqrt{1+{{e}^{x}}}+1}\]

If \[\int{\frac{x{{e}^{x}}}{\sqrt{1+{{e}^{x}}}}dx=f(x)\sqrt{1+{{e}^{x}

1.	If \[\int{\frac{x{{e}^{x}}}{\sqrt{1+{{e}^{x}}}}dx=f(x)\sqrt{1+{{e}^{x}}}-2\log \,\,g(x)+C,}\] then
A.	\[f(x)=x-1\]
B.	\[g(x)=\frac{\sqrt{1+{{e}^{x}}}-1}{\sqrt{1+{{e}^{x}}}+1}\]
C.	\[g(x)=\frac{\sqrt{1+{{e}^{x}}}+1}{\sqrt{1+{{e}^{x}}}-1}\]
D.	\[f(x)=2(2-x)\]
Answer» C. \[g(x)=\frac{\sqrt{1+{{e}^{x}}}+1}{\sqrt{1+{{e}^{x}}}-1}\]

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If \[\int{\frac{x{{e}^{x}}}{\sqrt{1+{{e}^{x}}}}dx=f(x)\sqrt{1+{{e}^{x}}}-2\log \,\,g(x)+C,}\] then

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