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#