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Question
Evaluate `int_1^"e" 1/(x(1 + log x)^2) "d"x`
Solution
Let I = `int_1^"e" 1/(x(1 + log x)^2) "d"x`
Put 1 + log x = t
∴ `1/x "d"x` = dt
When x = 1, t = 1 + log 1 = 1 + 0 = 1
When x = e, t = 1 + log e = 1 + 1 = 2
∴ I = `int_1^2 "dt"/"t"^2`
= `int_1^2 "t"^(-2) "dt"`
= `[("t"^(-1))/(-1)]_1^2`
= `-[1/"t"]_1^2`
= `-(1/2 - 1)`
∴ I = `-((-1)/2)`
= `1/2`
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