A) \[\sqrt{x-1}\]
B) \[\sqrt{x+1}\]
C) \[\sqrt{{{x}^{2}}+1}\]
D) \[\frac{x}{\sqrt{1+{{x}^{2}}}}\]
Correct Answer: D
Solution :
\[\int_{1}^{b}{f(x)\,dx=\sqrt{{{b}^{2}}+1}-\sqrt{2}}=\sqrt{{{b}^{2}}+1}-\sqrt{1+1}=[\sqrt{{{x}^{2}}+1}]_{1}^{b}\] \[\therefore \] \[f(x)=\frac{d}{dx}\sqrt{{{x}^{2}}+1}\]\[=\frac{2x}{2\sqrt{{{x}^{2}}+1}}=\frac{x}{\sqrt{{{x}^{2}}+1}}\].You need to login to perform this action.
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