How do you integrate $\int{\dfrac{1}{{{x}^{3}}}dx}$?

Hint: We first explain the term $\dfrac{dy}{dx}$ where $y=f\left( x \right)$. We then need to integrate the equation$\int{\dfrac{1}{{{x}^{3}}}dx}$ once to find all the solutions of the differential equation. We take one constant for the integration and use the formula of $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$.

Complete answer:
We have to find the integral of the equation $\dfrac{1}{{{x}^{3}}}$. The mathematical form is $\int{\dfrac{1}{{{x}^{3}}}dx}$.
The main function is $y=f\left( x \right)$. We can convert using index form as $\dfrac{1}{{{x}^{3}}}={{x}^{-3}}$
So, $\int{\dfrac{1}{{{x}^{3}}}dx}=\int{{{x}^{-3}}dx}$
We know the integral form of $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$. We put the value of $n=-3$.
Constant terms get separated from the integral.
Simplifying the integral form, we get \[\int{\dfrac{1}{{{x}^{3}}}dx}=\int{{{x}^{-3}}dx}=\dfrac{{{x}^{-3+1}}}{-3+1}+c=\dfrac{{{x}^{-2}}}{-2}+c\].
Here $c$ is another constant. The simplified form is $\int{\dfrac{1}{{{x}^{3}}}dx}=-\dfrac{1}{2{{x}^{2}}}+c$

Note:
The solution of the power equation integral is the equation of a power function. The formula is valid for all $n$ except $-1$. The first order differentiation of $-\dfrac{1}{2{{x}^{2}}}+c$ gives the tangent of the circle for a certain point which is equal to $\dfrac{dy}{dx}=\dfrac{1}{{{x}^{3}}}$.