What is #int_(0)^(1) 1/x^2dx #?

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Answer 1 · 2016-01-28T02:41:32

#int_0^1 1/x^2dx -> oo# and therefore is divergent.

In general, #intx^ndx = x^(n+1)/(n+1) + C#

Thus #int1/x^2dx = intx^(-2)dx = x^(-1)/(-1)+C = -1/x+C#

Now, as #1/x^2# is not continuous at #x=0#, we will need to set up an improper integral to continue.

#int_0^1 1/x^2dx = lim_(A->0^+)int_A^1 1/x^2dx#

#=lim_(A->0^+)[-1/x]_A^1#

#=lim_(A->0^+)1/A-1#

#->oo#

meaning the integral is divergent.