What is the antiderivative of $1/ x^2$ ?

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Answer 1 · 2016-09-10T11:20:27

$= - 1/x+C$

$int x^n dx = x^(n+1)/(n+1)+C" , where "n!=-1$

here, $n = -2$

$int x^(-2) dx$

$= x^(-2+1)/(-2+1)+C$

$=x^(-1)/(-1)+C$

$= - 1/x+C$

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