Question #07f78

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Answer 1 · 2016-08-12T19:50:54

Does not converge

Does it converge? We'll find out

#int_3^oo1/(2x-1)dx#

#= lim_{alpha to oo} int_3^alpha \ 1/(2x-1)dx#

#= lim_{alpha to oo} int_3^alpha \ 1/2 d/dx ln(2x-1)dx#

#= 1/2 lim_{alpha to oo} [ ln(2x-1) ]_3^alpha#

#= 1/2 lim_{alpha to oo} ln((2alpha-1)/3)#

#beta = (2alpha-1)/3#

#= 1/2 lim_{beta to oo} ln beta#

#= oo#