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Integral of x^5*ln(3x)dx
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\(u=\ln(3x)\) and \(dv=x^5\,dx\) Then \(du=dx/x and v=x^6/6\) It follows your integral equals \(=x^6\ln(3x)/6-\int x^5/6\, dx\) \(=x^6\ln(3x)/6-x^6/36 +C\)
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