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∫logx(1+logx)2 dx

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Question

`int logx/(1 + logx)^2  "d"x`

Sum
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Solution

Let I = `int logx/(1 + logx)^2  "d"x`

Put log x = t

∴ x = et

∴ dx = et dt

∴ I = `int "t"/(1 + "t")^2 "e"^"t" "dt"`

= `int "e"^"t" [(("t" + 1) - 1)/(1 + "t")^2] "dt"`

= `int "e"^"t" [("t" + 1)/(1 + "t")^2 - 1/(1 + "t")^2] "dt"`

= `int "e"^"t" [1/(1 + "t") - 1/(1 + "t")^2] "dt"`

Put f(t) = `1/(1 + "t")`

∴ f'(t) = `(-1)/(1 + "t")^2`

∴ I = `int "e"^"t" ["f"("t") + "f'"("t")] "dt"`

= et f(t) + c

= `"e"^"t"* 1/(1 + "t") + "c"`

∴ I = `x/(1 + logx) + "c"`

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Chapter 1.5: Integration - Q.5

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