Question

If $f\left(x\right)={\int }_1^x\frac{\ln t}{1+t}dt$ where $x>0$, then the value(s) of $x$ satisfying the equation, $f\left(x\right)+f\left(1/x\right)=2$ is

• A. $2$
• B. $e$
• C. $e^{-2}$
• D. $e^2$

Answer 1

Answer: (C), (D)

The correct options are
C $e^2$
D $e^{-2}$

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

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

Substituting $t=\frac{1}{u}\Rightarrow dt=(-\frac{1}{u^2})du$

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

$={\int }_1^x\frac{\ln u}{u(u+1)}du={\int }_1^x\frac{\ln t}{t(1+t)}dt$

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

$={\int }_1^x\frac{(1+t)\ln t}{t(1+t)}dt={\int }_1^x\frac{\ln t}{t}dt=\frac{1}{2}{\left(\ln x\right)}^2$

Hence $f\left(x\right)+f\left(\frac{1}{x}\right)=2\Rightarrow x=e^{\pm 2}$