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