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What is #int_1^oo 1/(1+x )-1/xdx#?

1 Answer

Jul 16, 2016

# = ln (1/2)#

Explanation:

#int_1^oo 1/(1+x )-1/xdx#

#= [ ln(1+x )- ln x ]_1^oo#

#= [ ln((1+x )/x) ]_1^oo#

#= [ ln((1/x+1 )/1) ]_1^oo#

#= lim_{x to oo} ln((1/x+1 )/1) - ln((1/1+1 )/1) #

#= lim_{x to oo} ln((1/x+1 )/1) - ln((1/1+1 )/1) #

#= ln 1 - ln 2 = ln (1/2)#

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