What is #int_1^e ln(2x)dx#?

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Answer 1 · 2015-11-14T20:45:03

#=(e-1)ln2+1#

We will need to make use of the standard form :
#intln u du=u lnu - u +C#

So use substitution and let #u=2x#. Then #du =2dx=>1/2du=dx#.

Limits : #x=1=>u=2and x=e=>u=2e#

Therefore the original integral becomes :

#1/2int_2^(2e)lnu du#

#=1/2[u ln u - u]_2^(2e)#

#=1/2[(2eln(2e)-2e)-(2ln2-2)]#

#=(e-1)ln2+1#