A) \[\frac{1}{(x-\sqrt{1+{{x}^{2}}})}+c\]
B) \[-\frac{1}{\sqrt{1+{{x}^{2}}}}+c\]
C) \[-\sqrt{1+{{x}^{2}}}+c\]
D) \[\text{ln}\,(x-\sqrt{1+{{x}^{2}}})+c\]
Correct Answer: B
Solution :
[b] Given that \[f(x)=ln\,\,(x-\sqrt{1+{{x}^{2}}})\] \[\int{f''(x)dx=f'(x)+c}\] where c is a constant \[=\frac{1}{(x-\sqrt{1+{{x}^{2}}})}.\left( 1-\frac{2x}{2\sqrt{1+{{x}^{2}}}} \right)+c\] \[=\frac{-(x-\sqrt{1+{{x}^{2}}})}{(\sqrt{1+{{x}^{2}}})(x-\sqrt{1+{{x}^{2}}})}+c=-\frac{1}{\sqrt{1+{{x}^{2}}}}+c\]
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