The integral 16 ∫ _ 1 ^ 2 dx x ^ 3 ( x ^ 2 + 2) ^ 2 is equal to

Given,

$I = 16{∫}_{1}^{2}\dfrac{dx}{{x}^{3}{\left({x}^{2} + 2\right)}^{2}}$

$⇒ I = 16{∫}_{1}^{2}\dfrac{dx}{{x}^{3}{x}^{4}{\left(1 + \dfrac{2}{{x}^{2}}\right)}^{2}}$

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

Then, $I = - 4{∫}_{3}^{\dfrac{3}{2}}\dfrac{dt}{{\left(\dfrac{2}{t - 1}\right)}^{2}{t}^{2}}$

$⇒ I = - 4{∫}_{3}^{\dfrac{3}{2}}{\left(\dfrac{t - 1}{2}\right)}^{2}\dfrac{dt}{{t}^{2}}$

$⇒ I = - \dfrac{4}{4}{∫}_{3}^{\dfrac{3}{2}}\left(1 - \dfrac{2}{t} + \dfrac{1}{{t}^{2}}\right)dt$

$⇒ I = - 1{\left[t - 2\ln \left|t\right| - \dfrac{1}{t}\right]}_{3}^{\dfrac{2}{2}}$

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

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

$⇒ I = \dfrac{11}{6} - \ln 4$