The integral $16{\int }_1^2\frac{dx}{x^3{\left(x^2+2\right)}^2}$ is equal to

Given,

$I=16{\int }_1^2\frac{dx}{x^3{\left(x^2+2\right)}^2}$

$\Rightarrow I=16{\int }_1^2\frac{dx}{x^3{x}^4{\left(1+\frac{2}{x^2}\right)}^2}$

Now let $1+\frac{2}{x^2}=t\Rightarrow \frac{-4}{x^3}dx=dt$

Then, $I=-4{\int }_3^{\frac{3}{2}}\frac{dt}{{\left(\frac{2}{t-1}\right)}^2{t}^2}$

$\Rightarrow I=-4{\int }_3^{\frac{3}{2}}{\left(\frac{t-1}{2}\right)}^2\frac{dt}{t^2}$

$\Rightarrow I=-\frac{4}{4}{\int }_3^{\frac{3}{2}}\left(1-\frac{2}{t}+\frac{1}{t^2}\right)\mathrm{dt}$

$\Rightarrow I=-1{\left[\mathrm{t}-2\ln \left|t\right|-\frac{1}{t}\right]}_3^{\frac{2}{2}}$

$\Rightarrow I=-1\left[\left(\frac{3}{2}-2\ln \frac{3}{2}-\frac{2}{3}\right)-\left(3-2\ln 3-\frac{1}{3}\right)\right]$

$\Rightarrow I=-1\left[2\ln 2-\frac{11}{6}\right]$

$\Rightarrow I=\frac{11}{6}-\ln 4$