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प्रश्न
Integrate the following with respect to x.
`1/(x(log x)^2`
बेरीज
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उत्तर
= `int 1/(x(log x)^2 "d"x`
= `int 1/("t")^2 "dt"`
`int "t"^-2 "dt" ["t"^(-2 + 1)/(-2 + 1)] + "c"`
= `- 1/"t" + "c"`
= `- 1/(log|x|) + "c"`
Let t = log x
`"dt"/("d"x) = 1/x`
dt = `1/x "d"x`
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