Double antiderivative problem with natural logs

Question

Is what I did here correct?

I'm trying to solve for $f(x)$ given $f''(x).$

$$f''(x) = \frac{1}{x^2} = x^{-2}$$

$$f'(x) = -x^{-1} = \frac{-1}{x}$$

$$f(x) = - \ln{x}$$

I guess the rules for antiderivation are for every $x$ to an exponent except for when the exponent is $-1?$

Answer 1

You missed the integration constant.

$$f'(x) = -x^{-1} = \frac{-1}{x}+C$$ $$f(x) = -\ln x + Cx + D$$

Answer 2

Yes it is, indeed for every real number $a\neq -1$

$$\int x^a\,dx=\frac{x^{a+1}}{a+1}(+c)$$

Answer 3

It will get more clear if you use the notation of differential dx:

f ''(x) = $\frac{d^2}{dx^2}$( f(x) ) = $\frac{d}{dx}$ ( $\frac{df(x)}{dx}$ ) = x$^{-2}$

$\iff$ d ( $\frac{df(x)}{dx}$ ) = x$^{-2}$ dx

$\iff$ $\int$ d ( $\frac{df(x)}{dx}$ ) = $\int$ x$^{-2}$ dx

$\iff$ $\frac{df(x)}{dx}$ = $\frac{-1}{x}$ + C

$\iff$ d f( x ) = ($\frac{-1}{x}$ + C)dx

$\iff$ $\int$ d(f(x)) = $\int$ ($\frac{-1}{x}$ + C)dx = f(x)

$\iff$ f(x) = $\int$ $\frac{-1}{x}$ dx + $\int$ C dx = -ln(x) + Cx + D