${\int }_2^3 \frac{2{\mathrm{x}}^2}{{\mathrm{x}}^4+3{\mathrm{x}}^2+1}\mathrm{dx}$ is equal to

Let,

$\begin{array}{l}\mathrm{I}={\int }_2^3 \left(\frac{2{\mathrm{x}}^2}{{\mathrm{x}}^4+3{\mathrm{x}}^2+1}\right)\mathrm{dx}\\ ={\int }_2^3 \frac{\left({\mathrm{x}}^2+1\right)}{{\mathrm{x}}^4+3{\mathrm{x}}^2+1}\mathrm{dx}+{\int }_2^3 \frac{\left({\mathrm{x}}^2-1\right)\mathrm{dx}}{{\mathrm{x}}^4+3{\mathrm{x}}^2+1}\\ ={\int }_2^3 \frac{\left[1+\left(1/{\mathrm{x}}^2\right)\right]\mathrm{dx}}{[\mathrm{x}-(1/\mathrm{x}){]}^2+5}+{\int }_2^3 \frac{\left[1-\left(1/{\mathrm{x}}^2\right)\right]\mathrm{dx}}{[\mathrm{x}+(1/\mathrm{x}){]}^2+1}\end{array}$

In 1st put $\mathrm{x}-\frac{1}{\mathrm{x}}=\mathrm{t},$ in 2nd  put $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}$

 $\begin{array}{l}\mathrm{I}={\int }_{3/2}^{8/3} \frac{\mathrm{dt}}{{\mathrm{t}}^2+5}+{\int }_{5/2}^{10/3} \frac{\mathrm{dy}}{{\mathrm{y}}^2+1}\\ =\frac{1}{\sqrt{5}}\left[{\tan }^{-1}\left(\frac{8}{3\sqrt{5}}\right)-{\tan }^{-1}\left(\frac{3}{2\sqrt{5}}\right)\right]+{\tan }^{-1}\left(\frac{10}{3}\right)-{\tan }^{-1}\left(\frac{5}{2}\right)\\ =\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{7\sqrt{5}}{54}\right)+{\tan }^{-1}\left(\frac{5}{56}\right)\end{array}$