How do you evaluate the integral #int lnxdx# from [1,e]?

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Answer 1 · 2016-12-27T06:56:16

The answer is #=1#

#intuv'dx=uv-intu'vdx#

Here,

#u=lnx#, #=>#, #u'=1/x#

#v'=1#, #=>#, #v=x#

Therefore,

#int_1 ^elnx=[xlnx]_1^e-int_1 ^e1/x*xdx #

#=[xlnx]_1^e-[x]_1^e#

#=e-0-e+1#

#=1#