Question

Let $F\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right)$, where $f\left(x\right)={\int }_l^x\frac{logt}{l+t}dt$. Then $F\left(e\right)$ equals

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

Answer 1

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

$\because f\left(x\right)={\int }_1^x\frac{x\log t}{1+t}dt$ and $F\left(e\right)=f\left(e\right)+f\left(\frac{1}{e}\right)$
$\Rightarrow F\left(e\right)={\int }_1^e\frac{\log t}{1+t}dt+{\int }_1^{1/e}\frac{x\log t}{1+t}dt$
$={\int }_1^e\frac{\log t}{1+t}dt+{\int }_1^e\frac{\log t}{t(1+t)}dt={\int }_1^e\frac{\log t}{t}dt$
$={\left[\frac{{\left(\log t\right)}^2}{2}\right]}_1^e=\frac{1}{2}[{\left(\log e\right)}^2-{\left(\log 1\right)}^2]=\frac{1}{2}$