Let I ( x ) = ( x + 1 ) x ( 1 + x e x ) 2 d x , ? x 0 . If lim x I ( x ) = 0 , then I ( 1 ) is equal to [2023]

Question

$Let I (x) = \int \frac{(x + 1)}{x {(1 + x e^{x})}^{2}} d x, x > 0 . If \lim_{x \to \infty} I (x) = 0, then I (1) is equal to$ [2023]

Smart Achievers · Accepted Answer

(2)

$We have,$ $I (x) = \int \frac{(x + 1)}{x {(1 + x e^{x})}^{2}} d x$

$= \int \frac{e^{x} (1 + x)}{x e^{x} {(1 + x e^{x})}^{2}} d x$

$Put 1 + x e^{x} = t, we get$

$\Rightarrow (x e^{x} + e^{x}) d x = d t \Rightarrow (1 + x) e^{x} d x = d t$

$∴$ $I (x) = \int \frac{d t}{(t - 1) t^{2}} = \int (- \frac{1}{t} - \frac{1}{t^{2}} + \frac{1}{t - 1}) d t$

$= - \log t + \frac{1}{t} + \log (t - 1) + c$

$= - \log (1 + x e^{x}) + \frac{1}{1 + x e^{x}} + \log (x e^{x}) + c$

$\Rightarrow I (x) = \frac{1}{1 + x e^{x}} + \log (\frac{x e^{x}}{1 + x e^{x}}) + c$

$Now, \lim_{x \to \infty} I (x) = 0 \Rightarrow c = 0$

$∴$ $I (1) = \frac{1}{1 + e} + \log (\frac{e}{1 + e})$

$= \frac{1}{1 + e} + 1 - \log (1 + e) \Rightarrow \frac{2 + e}{1 + e} - \log (1 + e)$

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