Mathematics

# Let $\displaystyle F\left ( x \right )=f\left ( x \right )+f\left ( \frac{1}{x} \right )$ where $\displaystyle f\left ( x \right )=\int_{1}^{x}\frac{\log t}{1+t}dt$ Then $F(e)$ is equal to?

$1/2$

##### SOLUTION
$\displaystyle F\left ( x \right )=\int_{1}^{x}\frac{\ln t}{1+t}dt+\int_{1}^{1/x}\frac{\ln t}{1+t}dt$
$\displaystyle F\left ( x \right )=\int_{1}^{x}\left ( \frac{\ln t}{1+t}+\frac{\ln t}{\left ( 1+t \right )t} \right )dt=\int_{1}^{t}\frac{\ln t}{t}dt=\frac{1}{2}\left ( \ln x \right )^{2}$
$F\left ( e \right )=1/2$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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