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 :

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

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

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

$=(e-1)ln2+1$

What is int_1^e ln(2x)dxint_1^e ln(2x)dx?