Answer
\[ - \frac{2}{{{x^2}}} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{4}{{{x^3}}}dx} \hfill \\
Write\,\,\frac{1}{{{x^3}}}\,\,as\,\,{x^{ - 3}} \hfill \\
\int_{}^{} {4{x^{ - 3}}dx} \hfill \\
Use\,\,the\,\,power\,\,rule \hfill \\
\int_{}^{} {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \hfill \\
Then \hfill \\
\int_{}^{} {4{x^{ - 3}}dx} = 4\,\left( {\frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right) + C \hfill \\
Simplifying \hfill \\
4\,\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} \right) + C \hfill \\
- 2{x^{ - 2}} + C \hfill \\
- \frac{2}{{{x^2}}} + C \hfill \\
\end{gathered} \]
