Answer
$f(x)=-\ln |x|+Cx+D$
Work Step by Step
$\left[\begin{array}{cc}
\text{}f'' & \text{particular antiderivative,}f'\\
1/x^{2}=x^{-2} & \dfrac{x^{-2+1}}{-2+1}=-x^{-1}
\end{array}\right]$
$f'(x)=-\displaystyle \frac{1}{x}+C$
$\left[\begin{array}{cc}
\text{}f' & \text{particular antiderivative,}f\\
-1/x=(-1)\cdot\dfrac{1}{x} & -\ln |x|\\
C & Cx
\end{array}\right]$
$f(x)=-\ln |x|+Cx+D$
