If $I_1=\int\limits_x^1 \frac{1}{1+t^2} d t$ and $I_2=\int\limits_1^{1 / x} \frac{1}{1+t^2} d t$ for $x>0$, then

$I_1=I_2$

Putting $t=\frac{1}{u}$ in $I_1$, we get

$I_1=\int\limits_{1 / x}^1 \frac{1}{1+\frac{1}{u^2}}\left(-\frac{1}{u^2}\right) d u=-\int\limits_{1 / x}^1 \frac{d u}{u^2+1}=\int\limits_1^{1 / x} \frac{1}{1+u^2} d u=I_2$