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) \[f(x)=2(2-x)\]
Correct Answer: B
Solution :
[b] \[I=\int{\frac{x{{e}^{x}}}{\sqrt{1+{{e}^{x}}}}dx}\] we have \[\int{\frac{{{e}^{x}}}{\sqrt{1+{{e}^{x}}}}dx=2\sqrt{1+{{e}^{x}}}}\] Integrating I by parts with x as first function \[I=x.2\sqrt{1+{{e}^{x}}}-\int{2\sqrt{1+{{e}^{x}}}dx}\] \[=2x\sqrt{1+{{e}^{x}}}-2\int{t\cdot \frac{2tdt}{{{t}^{2}}-1}(Putting\,\,1+{{e}^{x}}={{t}^{2}})}\] \[=2x\sqrt{1+{{e}^{x}}}-4\int{\frac{{{t}^{2}}-1+1}{{{t}^{2}}-1}dt}\] \[=2x\sqrt{1+{{e}^{x}}}-4\left[ t+\frac{1}{2}\log \frac{t-1}{t+1} \right]+c\] \[=2(x-2)\sqrt{1+{{e}^{x}}}-2\,\,\log \left( \frac{\sqrt{1+{{e}^{x}}}-1}{\sqrt{1+{{e}^{x}}}+1} \right)+c\]
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