How do you integrate $\dfrac{1}{{{x^2}}}dx?$

Hint:As we know that the definition of integration says that it is a process of finding functions whose derivative is given is named as anti differentiation or integration. We will solve the above question by the power rule of integration. The formula is $\int {{x^a}dx = \dfrac{{{x^{a + 1}}}}{{a + 1}},a \ne - 1} $. It will allow us to integrate any function that can be written as power of $x$.

Complete step by step answer:
Here we have to integrate $\int {\dfrac{1}{{{x^2}}}dx} $. We will first apply the exponent rule. It says that if we have $\dfrac{1}{{{a^b}}}$, then it can be written as ${a^{ - b}}$. So by applying this exponent rule, we can write $\dfrac{1}{{{x^2}}}$ as ${x^{ - 2}}$, since we have $a = x\,,\,b = 2$.

So now we have to integrate the new expression i.e. $\int {{x^{ - 2}}dx} $.
We will apply the power rule $\int {{x^a}dx = \dfrac{{{x^{a + 1}}}}{{a + 1}},a \ne - 1} $.
On comparing here we have $a = - 2$.Putting this in the formula we can write,
$\dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}} = \dfrac{{{x^{ - 1}}}}{{ - 1}} + C$.
We can write it as $ - {x^{ - 1}} + C$. Now we know that ${m^{ - 1}}$ can be written as $\dfrac{1}{m}$.So it can be written as $ - \dfrac{1}{x} + C$.

Hence the integration of $\dfrac{1}{{{x^2}}}dx=- \dfrac{1}{x} + C$.

Note:We should note that the power rule for integration is applied when the function is in numerator either with positive or negative power. In the above question we have the positive power i.e. $2$. Also we should not forget to write $ + C$ at the end of the solution. It is also known as the constant of the integration or arbitrary constant.