Answer
$\dfrac{-1}{x}-\dfrac{1}{2x^2}+C$
Work Step by Step
Given: $\int x^{-3}(x+1) dx$
Since, we have $\int x^n dx=\dfrac{x^{n+1}}{n+1}+C$
Thus, $\int x^{-3}(x+1) dx=\int (x^{-2}+x^{-3})dx=-x^{-1}-\dfrac{x^{-2}}{2}+C=\dfrac{-1}{x}-\dfrac{1}{2x^2}+C$
