How do you integrate #int ln(x)/x dx# using integration by parts?

1 Answer
Topscooter
Dec 20, 2015

#intln(x)/xdx = ln(x)^2/4#

Explanation:

Integration by parts is a bad idea here, you will constantly have #intln(x)/xdx# somewhere. It is better to change the variable here because we know that the derivative of #ln(x)# is #1/x#.

We say that #u(x) = ln(x)#, it implies that #du = 1/xdx#. We now have to integrate #intudu#.

#intudu = u^2/2# so #intln(x)/xdx = ln(x)^2/2#

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