Answer
The power rule was incorrectly applied.
Correct answer:$\displaystyle \quad-\frac{1}{2x^{2}}+C$
Work Step by Step
$\displaystyle \int\frac{1}{x}dx=\ln|x|+C$
(special form of the power rule, applied only when $n=-1)$
Here it is incorrectly applied, because $n=-3.$
$\displaystyle \int\frac{1}{x^3}dx=\int x^{-3}dx=\qquad $
The power rule applies
$\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ ,$\quad n\neq-1$
$=\displaystyle \frac{x^{-3+1}}{-3+1}+C$
$=\displaystyle \frac{x^{-2}}{-2}+C$
$=-\displaystyle \frac{1}{2x^{2}}+C$
