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प्रश्न
`int logx/(1 + logx)^2 "d"x`
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उत्तर
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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