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Question:

What is the integral of log (x^x)

Answer:

Let I = ∫log(x^x )dx

=> I = ∫x*logx dx

=> I = (x² *logx)/2 - ∫{(1/x)*x² /2}dx

=> I = (x² *logx)/2 - ∫{x/2}dx

=> I = (x² *logx)/2 - (1/2)*∫xdx

=> I = (x² *logx)/2 - (1/2)*(x² /2) + C

=> I = (x² *logx)/2 - x² /4 + C

So, ∫log(x^x )dx = (x² *logx)/2 - x² /4 + C

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