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?


ANSWER

$$1/2$$


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
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