Question

Let $F\left(x\right)=f\left(x\right)+f\left(\frac{1}{x}\right),$ where $f\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t}dt$.
Then the value of $F\left(e\right)$ is:

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

Answer 1

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

$F\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t}dt+\underset{1}{\overset{1/x}{\int }}\frac{\log t}{1+t}dt$
In the 2nd integral let $t=1/y$
$\Rightarrow dt=-\frac{1}{y^2}dy$
$\Rightarrow F\left(x\right)=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t}dt+\underset{1}{\overset{x}{\int }}\frac{-\log y}{1+\frac{1}{y}}(-\frac{dy}{y^2})$
$=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t}dt+\underset{1}{\overset{x}{\int }}\frac{\log y}{y(1+y)}dy$
$=\underset{1}{\overset{x}{\int }}\frac{\log t}{1+t}dt+\underset{1}{\overset{x}{\int }}\frac{\log t}{t(1+t)}dt$
$=\underset{1}{\overset{x}{\int }}\frac{\log t}{t}dt=\frac{1}{2}(\log x{)}^2$
$\therefore F\left(e\right)=\frac{1}{2}$