determine the function f if f''(x)=2/x^2, f(1)=1, f'(x)=1, x>0?

Question

Need all steps please.

Answer 1

By "f'(x)=1", I think that you mean "f'(1)=1", at least "x" must be changed to some number.
f''(x)=2/x^2
f'(x) = [integral sign]2dx /x^2
= -2/x + C_1
C_1 is determined to be 3 based on f'(1)=1
Thus f'(x) = -2/x + 3
f(x) = [integral sign](-2/x + 3) dx
= -2 Ln x + 3x + C_2
C_2 is determined to be -2 based on f(1)=1
Thus f(x) = -2 Ln x + 3x - 2