Question 1031670
integrate x^2 ln(x^3) dx
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substitute u = x^3 and  du = 3 x^2  dx =  1/3 integral ln(u) du
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For the integrand ln(u), integrate by parts,  integral f dg = f g- integral g df, where f = ln(u),     dg =   du,
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df = 1/u  du,     g = u
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df  =  1/3 u ln(u)-1/3 integral 1 du
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The integral of 1 is u:
  =  1/3 u ln(u)-u/3+constant
Substitute back for u = x^3:
  =  1/3 x^3 ln(x^3)-x^3/3+constant
:
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y' =  1/3 x^3 (ln(x^3)-1)+constant
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we can evaluate the definite integral and forget about the constant
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1/3(1)^3 ((ln(1^3)-1) - 0 =
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(ln(1) / 3) - 1/3 = 1/3
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note that area is positive
note that ln(1) = 0
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