Question

For $x>0,$ if $f(x)=\underset{1}{\overset{x}{\int }}\frac{{\log }_{e}t}{(1+t)}dt,$ then $f(e)+f\left(\frac{1}{e}\right)$ is equal to

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

Answer 1

Answer: (C)

$f(x)=\underset{1}{\overset{x}{\int }}\frac{{\log }_{e}t}{(1+t)}dt$
$f\left(\frac{1}{x}\right)=\underset{1}{\overset{1/x}{\int }}\frac{\ell nt}{1+t}dt,$ let $t=\frac{1}{y}$
$=+\underset{1}{\overset{x}{\int }}\frac{\ln y}{1+y}\cdot \frac{y}{y^2}dy$
$=\underset{1}{\overset{x}{\int }}\frac{\ell ny}{y(1+y)}dy$
hence
$f(x)+f\left(\frac{1}{x}\right)=\underset{1}{\overset{x}{\int }}\frac{(1+t)\ell nt}{t(1+t)}dt=\underset{1}{\overset{x}{\int }}\frac{\ell nt}{t}dt$
$=\frac{1}{2}{\ln }^2(x)$
so $f(e)+f\left(\frac{1}{e}\right)=\frac{1}{2}..(3)$