The value of the integral ${\int }_1^2 \left(\frac{{\mathrm{t}}^4+1}{{\mathrm{t}}^6+1}\right)\mathrm{dt}$ is

$\begin{array}{l}\mathrm{I}={\int }_1^2 \frac{{\mathrm{t}}^4+1}{{\mathrm{t}}^6+1}\mathrm{dt}={\int }_1^2 \frac{\left({\mathrm{t}}^4-{\mathrm{t}}^2+1+{\mathrm{t}}^2\right)}{\left({\mathrm{t}}^6+1\right)}\mathrm{dt}\\ ={\int }_1^2 \frac{{\mathrm{t}}^4-{\mathrm{t}}^2+1}{\left({\mathrm{t}}^6+1\right)}+\frac{\left({\mathrm{t}}^2\right)}{\left({\mathrm{t}}^6+1\right)}\mathrm{dt}={\int }_1^2 \frac{\mathrm{dt}}{{\mathrm{t}}^2+1}+\frac{1}{3}{\int }_1^2 \frac{3{\mathrm{t}}^2\mathrm{dt}}{{\left({\mathrm{t}}^3\right)}^2+1}\\ {\left.{\left.={\tan }^{-1}\mathrm{t}\right]}_1^2+\frac{1}{3}{\tan }^{-1}{\mathrm{t}}^3\right]}_1^2\\ =\left({\tan }^{-1}2-{\tan }^{-1}\left(1\right)\right)+\frac{1}{3}\left({\tan }^{-1}\left(8\right)-{\tan }^{-1}\left(1\right)\right)\\ ={\tan }^{-1}2+\frac{1}{3}{\tan }^{-1}8-\frac{\mathrm{\pi }}{3}\end{array}$