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 $f(x)+f\left(\frac{1}{x}\right)=\frac{\left(\log _e x\right)^2}{2}$

Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.

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.