Question

What is the antiderivative of \[\dfrac{1}{{{x^2}}}\]?

Answer 1

Hint: Here in this question given an indefinite integral, we have to find the integrated value of a given function. Here we use the standard formulas for the integration and on simplifying we obtain the required solution for the given function.

Complete step by step solution:
The antiderivative means the inverse of the differentiation. The inverse of the differentiation is an integration.
integration is the inverse process of differentiation. An integral which is not having any upper and lower limit is known as an indefinite integral.
Consider the given function.
\[ \Rightarrow \,\dfrac{1}{{{x^2}}}\]
Integrate with respect to x, then
\[ \Rightarrow \,\int {\dfrac{1}{{{x^2}}}\,} dx\]---------(1)
Now we take \[{x^2}\] to the numerator which is present in the denominator. While the term which is taken to the numerator from the denominator the power will be negative. It is written as
\[ \Rightarrow \,\int {{x^{ - 2}}\,} dx\]
By the standard formulas of integration we know that the \[\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\], using this the given integral is written as and the value of n will be -2. So we have
\[ \Rightarrow \dfrac{{{x^{ - 2 + 1}}}}{{ - 2 + 1}} + c\]
On simplifying we have
\[ \Rightarrow \dfrac{{{x^{ - 1}}}}{{ - 1}} + c\]
In the denominator the x term has power negative so we write that term in the denominator. While writing the term in the denominator it will be positive and hence we have
\[ \Rightarrow - \dfrac{1}{x} + c\], where c is the integration constant
Hence, it’s a required solution.
So, the correct answer is “\[ - \dfrac{1}{x} + c\]”.

Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.