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Find : ∫ B a Log X X Dx

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प्रश्न

Find : `∫_a^b logx/x` dx

योग
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उत्तर

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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2018-2019 (March) 65/3/3

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