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What is the antiderivative of #(1/(x^2))#?

1 Answer

Apr 18, 2016

#-1/x+C#

Explanation:

#d/dx(x^r+C) = rx^(r-1)# (for constant #C#)

So, for all #s != 0#, the antiderivative of #x^s# is #x^(s+1)/(s+1)+C#.

#1/x^2 = x^-2# so its antiderivative is #x^(-2+1)/(-2+1) = x^-1/-1 = -1/x# don't forget the #+C#.

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