Question

Let $F\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$ where $f\left(x\right)={\int }_1^x\frac{\log t}{1+t}dt$

Then $F\left(e\right)$ is equal to?

• A. $1$
• B. $2$
• C. $1/2$
• D. $0$

Answer 1

The correct option is C $1/2$

$F\left(x\right)={\int }_1^x\frac{\ln t}{1+t}dt+{\int }_1^{1/x}\frac{\ln t}{1+t}dt$
$F\left(x\right)={\int }_1^x(\frac{\ln t}{1+t}+\frac{\ln t}{(1+t)t})dt={\int }_1^t\frac{\ln t}{t}dt=\frac{1}{2}{\left(\ln x\right)}^2$
$F\left(e\right)=1/2$