How do you integrate #(ln(x)*(1/x))dx#?

Question

Emma Aldridge · Accepted Answer

# = 1/2 (ln x)^2 + C# #int dx qquad ln(x)*1/x# you can do this by inspection as #(ln x) ' = 1/x# so we can trial #alpha (ln x)^2 # as a solution. So #(alpha (ln x)^2 + C)' = 2 alpha ln x 1/x implies 2 alpha = 1, alpha = 1/2# if you don't fancy that you could use IBP : #int uv' = uv - int u'v# #u = ln x, u' = 1/x# #v' = 1/x, v = ln x# #implies color{red}{int dx qquad ln(x)1/x} = (ln x)^2 - color{red}{int dx qquad ln x 1/x }+ C# #implies 2 int dx qquad ln(x)1/x = (ln x)^2 + C# #implies int dx qquad ln(x)1/x = 1/2 (ln x)^2 + C#

William Abdalla · Answer

undefined To integrate ln(x)*(1/x) dx, you can use integration by parts. Let u = ln(x) and dv = (1/x) dx. Then, differentiate u to get du = (1/x) dx and integrate dv to get v = ln(x).

Applying the integration by parts formula:

∫u dv = uv - ∫v du

This yields:

∫ln(x)*(1/x) dx = ln(x) * ln(x) - ∫ln(x) * (1/x) dx

Now, the integral on the right side is the same as the original integral, so we can rewrite it as:

∫ln(x)*(1/x) dx = ln(x) * ln(x) - ∫ln(x)*(1/x) dx

Adding ∫ln(x)*(1/x) dx to both sides:

2∫ln(x)*(1/x) dx = ln(x) * ln(x)

Now, solve for ∫ln(x)*(1/x) dx:

∫ln(x)*(1/x) dx = (1/2) * ln(x) * ln(x) + C

Where C is the constant of integration.