# Evaluate the following integral. \int x^{-3}(x+1)dx

Evaluate the following integral.
$\int {x}^{-3}\left(x+1\right)dx$
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averes8
$\int {x}^{-3}\left(x+1\right)dx=\int \left({x}^{-2}+{x}^{-3}\right)dx$
$=\frac{{x}^{-2+1}}{-2+1}+\frac{{x}^{-3+1}}{-3+1}+c$
Hence, $\int {x}^{-3}\left(x+1\right)dx=-\frac{1}{x}-\frac{1}{2{x}^{2}}+c$