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Calculation: definite integral ∫ (upper 1, lower 0) ln
∫(0,1]ln xdx
=xlnx(0,1]-x(0,1]
=-1
The comparison of ∫ xdx and ∫ ln (x + 1) definite integrals in (0,1) is significant,
Let g (x) = x-ln (x + 1), G '(x) = 1-1 / (1 + x), when x is between 0,1, G' (x) > = 0, so g (x) on (0,1) increases, and G (0) = 0, so g (x) > = 0, that is, x > ln (1 + x), which is omitted
In fact, it can also be compared with the value obtained by the partial integration method
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