Question
If $$f\left( x \right) = \ln \left( {x - \sqrt {1 + {x^2}} } \right),$$ then what is $$\int {f''\left( x \right)dx} $$ equal to ?
A.
$$\frac{1}{{\left( {x - \sqrt {1 + {x^2}} } \right)}} + c$$
B.
$$ - \frac{1}{{\sqrt {1 + {x^2}} }} + c$$
C.
$$ - \sqrt {1 + {x^2}} + c$$
D.
$$\ln \left( {x - \sqrt {1 + {x^2}} } \right) + c$$
Answer :
$$ - \frac{1}{{\sqrt {1 + {x^2}} }} + c$$
Solution :
Given that $$f\left( x \right) = \ln \left( {x - \sqrt {1 + {x^2}} } \right)$$
$$\int {f''\left( x \right)dx} = f'\left( x \right) + c$$ where $$c$$ is a constant
$$\eqalign{
& = \frac{1}{{\left( {x - \sqrt {1 + {x^2}} } \right)}}.\left( {1 - \frac{{2x}}{{2\sqrt {1 + {x^2}} }}} \right) + c \cr
& = \frac{{ - \left( {x - \sqrt {1 + {x^2}} } \right)}}{{\left( {\sqrt {1 + {x^2}} } \right)\left( {x - \sqrt {1 + {x^2}} } \right)}} + c \cr
& = - \frac{1}{{\sqrt {1 + {x^2}} }} + c \cr} $$