Question

Evaluate `int_1^"e" 1/(x(1 + log x)^2)  "d"x`

Answer 1

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`