Let f(x)+f1x=F(x), where f(x)=∫1x ln⁡t1+tdt. then F(e) =

Fe=f(e)+f1e=∫1e ln⁡t1+tdt+∫11/e ln⁡t1+tdtSetting t=1/u in the second integral, F(e)=∫1e ln⁡t1+tdt+∫1e ln⁡uu(1+u)du=∫1e ln⁡t1+t1+1tdt=∫1e ln⁡ttdt=(ln⁡t)221e=12

Let f(x)+f1x=F(x), where f(x)=∫1x ln⁡t1+tdt. then F(e) =

Fe=f(e)+f1e=∫1e ln⁡t1+tdt+∫11/e ln⁡t1+tdtSetting t=1/u in the second integral, F(e)=∫1e ln⁡t1+tdt+∫1e ln⁡uu(1+u)du=∫1e ln⁡t1+t1+1tdt=∫1e ln⁡ttdt=(ln⁡t)221e=12

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Let $f (x) + f (\frac{1}{x}) = F (x)$ , where $f (x) = \int_{1}^{x} \frac{\ln t}{1 + t} dt$ . then F(e) =

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$0$

$2$

$1$

$\frac{1}{2}$

answer is D.

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Detailed Solution

$Fe = f (e) + f (\frac{1}{e}) = \int_{1}^{e} \frac{\ln t}{1 + t} dt + \int_{1}^{1 / e} \frac{\ln t}{1 + t} dt$
Setting t=1/u in the second integral, $F (e) = \int_{1}^{e} \frac{\ln t}{1 + t} dt + \int_{1}^{e} \frac{\ln u}{u (1 + u)} du$
$= \int_{1}^{e} \frac{\ln t}{1 + t} (1 + \frac{1}{t}) dt = \int_{1}^{e} \frac{\ln t}{t} dt = {\frac{(\ln t)^{2}}{2}|}_{1}^{e} = \frac{1}{2}$