$ I =\int\limits_1^2\left(\frac{ t ^4+1}{ t ^6+1}\right) dt $
$ =\int\limits_1^2 \frac{\left( t ^4+1- t ^2\right)+ t ^2}{\left( t ^2+1\right)\left( t ^4- t ^2+1\right)} dt $
$=\int\limits_1^2\left(\frac{1}{ t ^2+1}+\frac{ t ^2}{ t ^6+1}\right) dt $
$=\int\limits_1^2\left(\frac{1}{ t ^2+1}+\frac{1}{3} \frac{3 t ^2}{\left( t ^3\right)^2+1}\right) dt $
$ =\tan ^{-1}(t)+\left.\frac{1}{3} \tan ^{-1}\left(t^3\right)\right|_1 ^2 $
$ =\left(\tan ^{-1}(2)-\tan ^{-1}(1)\right)+\frac{1}{3}\left(\tan ^{-1}\left(2^3\right)-\tan ^{-1}\left(1^3\right)\right) $
$ =\tan ^{-1}(2)+\frac{1}{3} \tan ^{-1}(8)-\frac{\pi}{3}$