Question

f(x)=ln|x+√[1+x)^2| is even or odd

Answer 1

I think here you write in place of (1+x^2) is (1+x)^2

f (x ) = ln | x +√(1+x^2) |

put x = -x

f(-x ) = ln | -x +√(1+x^2) |

add f(x ) and f( -x )

f(x) + f(-x ) =ln |x+√(1+x)| +ln|x+√(1+x^2)|

=ln{(√(1+x^2))^2-x^2}

=ln(1+x^2 -x^2)

=ln(1) =0

hence ,
f(x)+f(-x) =0
so, f(x) is an odd function