What is the antiderivative of $\frac{1}{x^{2}}$ $1/ x^2$ ?

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What is the antiderivative of $\frac{1}{x^{2}}$ $1/ x^2$ ?

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Answer 1 · 2015-06-06T22:33:22

It's as easy as taking the negative exponent and integrate with the exponent rule.

The antiderivative is the same as the integral, and I guess you want to compute $\int \frac{1}{x^{2}} d x$ $int1/x^2dx$ . Since $\frac{1}{x^{2}} = x^{- 2}$ $1/x^2=x^(-2)$ , let's calculate the integral using the exponent algorythm:

$\int x^{- 2} d x = \frac{x^{- 1}}{- 1} + C = - \frac{1}{x} + C$ $intx^(-2)dx=x^(-1)/(-1)+C=-1/x+C$

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What is the antiderivative of $\frac{1}{x^{2}}$ $1/ x^2$ ?

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