Let $F(x)=f(x)+f\left(\frac{1}{x}\right)$, where $f(x)=\int_1^x \frac{\log t}{1+t} d t$. Then $F(e)$ equals

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Aicharya · Accepted Answer

$f(x)=\int_1^x \frac{\log t}{1+t} d t$ $F(e)=f(e)+f\left(\frac{1}{e}\right)$ $F(e)=\int_1^e \frac{\log t}{1+t} d t+\int_1^{1 / e} \frac{\log t}{1+t} d t$ $=\int_1^e \frac{\log t}{1+t}+\int_1^e \frac{\log t}{t(1+t)} d t$ $=\int_1^e \frac{\log t}{t} d t=\frac{1}{2}$.

Let $F(x)=f(x)+f\left(\frac{1}{x}\right)$, where $f(x)=\int_1^x \frac{\log t}{1+t} d t$. Then $F(e)$ equals

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