What is the integral of #int 1 /(2x)dx#?

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Answer 1 · 2018-04-03T05:17:22

#1/2timesInx +c#

#int1/(2x) times dx#
=#1/2timesint1/x dx#
#1/2timesInx +c#

Answer 2 · 2018-04-03T05:35:22

#1/2ln|x|+C#

We got: #int1/(2x) \ dx#

Taking out the constant, we get,

#=1/2int1/x \ dx#

Now, we have the common integral, where #int1/x \ dx=ln|x|+C#, and so,

#=1/2ln|x|+C#