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Question
Solve the following : `int_1^2 dx/(x(1 + logx)^2`
Sum
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Solution
Let I = `int_1^2 dx/(x(1 + logx)^2`
Put 1 + log x = t
∴ `(1)/x*dx` = dt
When x = 1, t = 1 + log 1
= 1 + 0 = 1
When x = 2, t = 1 + log 2
∴ I = `int_1^(1 + log2) "dt"/"t"^2`
= `[- 1/"t"]_1^(1 + log 2)`
= `-(1/(1 + log 2) - 1)`
= `-((1 - 1 - log 2)/(1 + log 2))`
∴ I = `log2/(1 + log2)`.
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