How do you integrate $\int{\left( 2x-3{{x}^{2}} \right)}dx$ ?

Hint: To solve these questions first separate the terms. Then apply the power rule of integration to the terms separately to get the answer. By integrating the terms separately, the question becomes easier to solve and simplify.

Complete step-by-step solution:
Integration, in general, can be defined as the summation of data that is discrete or separate.
We can use integration to find the functions such as area, volume, etc. of various functions under the graph.
Integration, in layman’s terms, can be defined as the method of adding or summing up the parts to find the whole of something.
Given:
$\int{(2x-3{{x}^{2}})}dx$ ...$(i)$
To integrate the above expression, we must know the integration formula
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\forall x\in R,x\ne -1$ ...$(ii)$
Also, we know that $\int{(f(x)+g(x))dx=\int{f(x)dx+\int{g(x)dx}}}$ …$(iii)$
Comparing equation $(i)$ and$(iii)$, then rewriting equation$(i)$, we get:
$\Rightarrow \int{(2x-3{{x}^{2}})}dx=\int{2xdx-\int{3{{x}^{2}}dx}}$
Now, applying the integration formula given in the equation$(ii)$, we get:
$\Rightarrow \int{2xdx-\int{3{{x}^{2}}dx}}=2\left( \dfrac{{{x}^{1+1}}}{1+1} \right)-3\left( \dfrac{{{x}^{2+1}}}{2+1} \right)+C$, where $C$is the constant of integration.
Further simplifying the above expression we get,
$\Rightarrow 2\left( \dfrac{{{x}^{1+1}}}{1+1} \right)-3\left( \dfrac{{{x}^{2+1}}}{2+1} \right)+C=2\left( \dfrac{{{x}^{2}}}{2} \right)-3\left( \dfrac{{{x}^{3}}}{3} \right)+C$
Cancelling the like terms from the numerator and denominator, we get:
$\Rightarrow 2\left( \dfrac{{{x}^{2}}}{2} \right)-3\left( \dfrac{{{x}^{3}}}{3} \right)+C={{x}^{2}}-{{x}^{3}}+C$, which is the required answer.
Hence, on integrating $\int{(2x-3{{x}^{2}})}dx$ by separating the terms and applying the power rule of integration we get the answer as ${{x}^{2}}-{{x}^{3}}+C$.

Note: Integration is also known as anti-derivative since it does the opposite of what derivation does. Integration has various applications in real life. It can be used to find the area between curves, the volume of solids, work done, surface area, center of mass, probability, and much more.
While solving these types of questions especially which involve the use of the formula$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}\forall x\in R,x\ne -1$, the restriction $x\ne -1$ is incredibly important.