If `(X) = f(X), f(0) = 1` then `int (dx)/(f(x) + f(-x))=`

Options

A. `log(e^(2x) +1) + C`
B. `log(e^(x) + e^(-x)) +c`
C. `tan^(-1) (e^(x))+ c`
D. None of these

Answer

3

Explanation

`f'(x) =f(x)`
`rArr(f'(x))/(f(x)) =1`
`int(f(x))/(f(x)) dx = int 1 dx`
`rArr log f(x) = x +C`
at`x=0,f(0)=1 ("given")`
`therefore log f(0) = 0 + C rArr C=0`
` therefore log f(x) = x rArr f(x) = e^(x)`
Now
`I = int (dx)/(f(x) + f(-x))`
`= int(dx)/( e^(x) +e^(-x))=int(e^(x))/(e^(2x) +1)dx`
`= tan^(-1) e^(x) +C`