Integral Calculus Question 141
Question: If $ f(x)=ln(x-\sqrt{1+x^{2}}) $ , then what is $ \int{f’’(x)dx} $ equal to?
Options:
A) $ \frac{1}{(x-\sqrt{1+x^{2}})}+c $
B) $ -\frac{1}{\sqrt{1+x^{2}}}+c $
C) $ -\sqrt{1+x^{2}}+c $
D) $ ln,(x-\sqrt{1+x^{2}})+c $
Show Answer
Answer:
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}})}.( 1-\frac{2x}{2\sqrt{1+x^{2}}} )+c $ $ =\frac{-(x-\sqrt{1+x^{2}})}{(\sqrt{1+x^{2}})(x-\sqrt{1+x^{2}})}+c=-\frac{1}{\sqrt{1+x^{2}}}+c $
BETA
