What is the integral of #ln(x)/x#?

1 Answer
Dec 15, 2014

Lets start by breaking down the function.

#(ln(x))/x = 1/x ln(x)#

So we have the two functions;

#f(x) = 1/x#
#g(x) = ln(x)#

But the derivative of #ln(x)# is #1/x#, so #f(x) = g'(x)#. This means we can use substitution to solve the original equation.

Let #u = ln(x)#.

#(du)/(dx) = 1/x#

#du = 1/x dx#

Now we can make some substitutions to the original integral.

#int ln(x) (1/x dx) = int u du = 1/2 u^2 + C#

Re-substituting for #u# gives us;

#1/2 ln(x)^2 +C#