A) 0
B) log 2
C) \[\log \frac{1}{2}\]
D) None of these
Correct Answer: A
Solution :
Let \[f(x)=\log (x+\sqrt{1+{{x}^{2}}})\] Now,\[f(-x)=\log \left( \sqrt{1+{{x}^{2}}}-x \right)=\log (\sqrt{1+{{x}^{2}}}-x).\frac{(\sqrt{1+{{x}^{2}}}+x)}{(\sqrt{1+{{x}^{2}}}+x)}\] \[=\log \frac{[(1+{{x}^{2}})-{{x}^{2}}]}{(\sqrt{1+{{x}^{2}}}+x)}\]\[=\log 1-\log (\sqrt{1+{{x}^{2}}}+x)\] \[=-\log (\sqrt{1+{{x}^{2}}}+x)\] \[=-f(x)\] Hence, \[\int_{\,-1}^{\,1}{\log \,(x+\sqrt{1+{{x}^{2}}})=0}\], \[\left[ \because \int_{\,-a}^{\,a}{f(x)=0,\,}\text{if }f(-x)=-f(x) \right]\].
You need to login to perform this action.
You will be redirected in
3 sec
