Question
If $$\int {\frac{{x{e^x}}}{{\sqrt {1 + {e^x}} }}} dx = f\left( x \right)\sqrt {1 + {e^x}} - 2\,\log \,g\left( x \right) + C,$$ then :
A.
$$f\left( x \right) = x - 1$$
B.
$$g\left( x \right) = \frac{{\sqrt {1 + {e^x}} - 1}}{{\sqrt {1 + {e^x}} + 1}}$$
C.
$$g\left( x \right) = \frac{{\sqrt {1 + {e^x}} + 1}}{{\sqrt {1 + {e^x}} - 1}}$$
D.
$$f\left( x \right) = 2\left( {2 - x} \right)$$
Answer :
$$g\left( x \right) = \frac{{\sqrt {1 + {e^x}} - 1}}{{\sqrt {1 + {e^x}} + 1}}$$
Solution :
$$\eqalign{
& I = \int {\frac{{x{e^x}}}{{\sqrt {1 + {e^x}} }}} dx{\text{ we have}} \cr
& \int {\frac{{{e^x}}}{{\sqrt {1 + {e^x}} }}} dx = 2\sqrt {1 + {e^x}} \cr} $$
Integrating $$I$$ by parts with $$x$$ as first function
$$\eqalign{
& I = x.2\sqrt {1 + {e^x}} - \int {2\sqrt {1 + {e^x}} } dx \cr
& = 2x\sqrt {1 + {e^x}} - 2\int {t.\frac{{2t\,dt}}{{{t^2} - 1}}\,\,\,\,\,\,} \left( {{\text{Putting }}1 + {e^x} = {t^2}} \right) \cr
& = 2x\sqrt {1 + {e^x}} - 4\int {\frac{{{t^2} - 1 + 1}}{{{t^2} - 1}}dt} \cr
& = 2x\sqrt {1 + {e^x}} - 4\left[ {t + \frac{1}{2}\log \frac{{t - 1}}{{t + 1}}} \right] + c \cr
& = 2\left( {x - 2} \right)\sqrt {1 + {e^x}} - 2\,\log \left( {\frac{{\sqrt {1 + {e^x}} - 1}}{{\sqrt {1 + {e^x}} + 1}}} \right) + c \cr} $$