How do I find the integral ∫(ln(x))^2dx ?

This problem is all about using integration by parts, so let's start by quoting the formula for integration by parts:

    ∫u*(dv/dx)dx = uv - ∫v*(du/dx)dx

To get the integral we want on the left hand side we can use the subtitutions u = dv/dx = ln(x). This means that we will have to find ∫ln(x)dx, this is also done using integration by parts:

To find ∫ln(x)dx we can use the substitutions u = ln(x) and dv/dx = 1. Using the formula above will then give us:

   ∫ln(x)*1dx = ln(x)*x - ∫x*(1/x)dx

 = xln(x) - ∫dx = xln(x) - x = x(ln(x)-1)

Using this we can now use our original substitutions in the formula to get:

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

 = xln(x)*(ln(x)-1) - ∫(ln(x)-1)dx 

 = xln(x)*(ln(x)-1) - x(ln(x)-1) + x + c 

Now we just have to tidy this up to get our final answer:

  ∫(ln(x))^2dx = x[(ln(x)+1)^2 + 1] + c

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2 years ago

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