Statement-1: If $f(x)=\int\limits_1^x \frac{\log _e t}{1+t+t^2} d t$, then $f(x)=f\left(\frac{1}{x}\right)$ for all $x>0$.
Statement-2: If $f(x)=\int\limits_1^x \frac{\log _e t}{1+t} d t$, then $f(x)+f\left(\frac{1}{x}\right)=\frac{\left(\log _e x\right)^2}{2}$
Answer & explanation
Correct answer: option 2
If $f(x)=\int\limits_1^x \frac{\log _e t}{1+t+t^2} d t$, then
$f\left(\frac{1}{x}\right) =\int\limits_1^{1 / x} \frac{\log _e t}{1+t+t^2} d t$
$\Rightarrow f\left(\frac{1}{x}\right) =\int\limits_1^x \frac{\log _e u}{1+u+u^2} d u$, where $t=\frac{1}{u}$
$\Rightarrow f\left(\frac{1}{x}\right)=f(x)$
So, statement -1 is true.
If $f(x)=\int\limits_1^x \frac{\log _e t}{2+t} d t$, then
$f\left(\frac{1}{x}\right)=\int\limits_1^{1 / x} \frac{\log _e t}{1+t} d t$
$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_1^x \frac{\log _e u}{u(1+u)} d u$, where $ t=\frac{1}{u}$
$\Rightarrow f\left(\frac{1}{x}\right)=\int\limits_1^x \frac{\log _e t}{t(1+t)} d t$
∴ $f(x)+f\left(\frac{1}{x}\right)=\int\limits_1^x \frac{\log _e t}{t+1}\left(1+\frac{1}{t}\right) d t$
$=\int\limits_1^x \frac{\log t}{t} d t=\left[\frac{\left(\log _e t\right)^2}{2}\right]_1^x=\frac{\left(\log _e x\right)^2}{2}$
So, statement-2 is also true.