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Find: ∫logx(1+logx)2dx

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Question

Find: `int logx/(1 + log x)^2 dx`

Sum
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Solution

`int logx/(1 + log x)^2 dx = int (log x + 1 - 1)/(1 + log x)^2 dx`

= `int 1/(1 + log x) dx - int 1/(1 + log x)^2 dx`

= `1/(1 + log x) xx x - int (-1)/(1 + log x)^2 xx 1/x xx xdx - int 1/(1 + log x)^2 dx`

= ` x/(1 + log x) + c`

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2021-2022 (March) Term 2 Sample

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