What is the Integration of $ dx $ ?

Hint: In order to determine the integration of $ dx $ write it as $ 1dx $ which we can further write as $ {x^0}dx $ . Now, Integrate the values obtained using the power rule that is $ \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1} $ and hence we get our result. Always add constant terms at last after integrating the equation.

Complete step by step solution:
We are given with the value $ dx $ .
First write the value of $ dx $ as $ 1dx $ .Since, we know that $ {x^0} = 1 $ , so write the same in the previous value and we get:
 $ dx = 1dx = {x^0}dx $
Now, Integrate the value and we get:
 $ \int {dx} = \int {1dx} = \int {{x^0}dx} $
Using the power rule which is $ \int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}},n \ne - 1} $ , we can solve the integration value further and hence, we get:
 $ \int {{x^0}dx = \dfrac{{{x^{0 + 1}}}}{{0 + 1}}} = \dfrac{{{x^1}}}{1} = x $
But this cannot be the accurate value because when we derive an equation its constant is removed, so let’s add $ C $ as constant value and we get:
\[\int {{x^0}dx} = x + C\]
Hence, the Integration of $ dx $ is $ x + C $
So, the correct answer is “\[ x + C\]”.

Note: It’s very important to add a constant after integrating the equation because when we derive an equation their constant is removed. For example, let’s see an equation: $ 2x + 2 $ Derivative this with respect to $ x $ and we get: $ \dfrac{{d(2x + 2)}}{{dx}} = \dfrac{{d2x}}{{dx}} + \dfrac{{d2}}{{dx}} = 2 + 0 = 2 $ .Because we know that derivation of constant term is zero. Now Let’s check and integrate the answer obtained and we get:
 $ \int {2dx} = 2x $ and we can see that the constant term is lost. That’s why we need to add a constant term at last after integration.