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Question
Find : `∫_a^b logx/x` dx
Sum
Solution
Put `log x = t ⇒ 1/x dx = dt`
⇒ `x = a ⇒ t = loga & x = b ⇒ t = log b`
`therefore I = ∫_log a ^log b t dt`
= `t^2/2|_log a^log b`
= `1/2 [(log b)^2 - (log a)^2]`
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Definite Integrals
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