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, $F\left(e\right)$ equals

• A. $\frac{1}{2}$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is A $\frac{1}{2}$

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

put $t=\frac{1}{p}⟹dt=-\frac{dp}{p^2}$

$f\left(x\right)={\int }_1^{1/x}\frac{\log p}{p(1+p)}dp$

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

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

$⟹F\left(e\right)=\frac{1}{2}$