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

Answer: (A)

$\frac{1}{2}$

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

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

Let $t=\frac{1}{4}$ $\therefore dt=\frac{-1}{u^2}du$

Also, $t=1\Rightarrow u=1$ $t=\frac{1}{x}\Rightarrow u=x$

$f\left(\frac{1}{x}\right)={\int }_1^x\frac{{\log }_e\left(\frac{1}{u}\right)}{1+\frac{1}{u}}\times \frac{-1}{u^2}du$

$\Rightarrow f\left(\frac{1}{x}\right)={\int }_1^x\frac{{\log }_{e}u}{(1+u)u}du={\int }_1^x\frac{{\log }_{e}t}{(1+t)t}dt\to \left(2\right)$

From $\left(1\right)and\left(2\right)$ we get,

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

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

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

$\Rightarrow f\left(x\right)+f\left(\frac{1}{x}\right)={\int }_0^{{\log }_{e}x}vdv,$ where $v={\log }_{e}t$ and $\frac{1}{t}dt=dv$

$\Rightarrow f\left(x\right)+f\left(\frac{1}{x}\right)={\left[\frac{v^2}{2}\right]}_0^{{\log }_{e}x}$

$=\frac{1}{2}{\left({\log }_{e}x\right)}^2$

$\Rightarrow f\left(e\right)+f\left(\frac{1}{e}\right)=\frac{{\left({\log }_{e}x\right)}^2}{2}=\frac{1}{2}.$

Hence, the answer is $\frac{1}{2}.$