How do you find the integral #ln x / x#?

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Answer 1 · 2015-09-12T19:08:24

#int ln x / x = (ln x)^2/2 + C#

#int ln x / x = ?#

This is a classic application of u -substitution.

Let #u = ln x#. Then #du = 1/x dx#.

Now we have

#int ln x/x dx = int u du#

We can evaluate this easily by using the power rule:

#int u du = u^2 /2 + C#

Now, substituting back #u#, we find

#u^2 /2 + C = (ln x)^2/2 + C#

And there is our solution.

#int ln x / x = (ln x)^2/2 + C#