The integral 16 1 2 d x x 3 ( x 2 + 2 ) 2 is equal to [2023]

Question

$The integral 16 \int_{1}^{2} \frac{d x}{x^{3} {(x^{2} + 2)}^{2}} is equal to$ [2023]

Smart Achievers · Accepted Answer

(1)

Let $I = 16 \int_{1}^{2} \frac{d x}{x^{3} {(x^{2} + 2)}^{2}} = 16 \int_{1}^{2} \frac{1}{x^{7} {(1 + \frac{2}{x^{2}})}^{2}} d x$

Let $1 + \frac{2}{x^{2}} = p \Rightarrow - \frac{4}{x^{3}} d x = d p \Rightarrow \frac{d x}{x^{3}} = \frac{- 1}{4} d p$

$I = - 4 \int_{3}^{3 / 2} \frac{d p}{\frac{4}{{(p - 1)}^{2}} \cdot p^{2}}$

$= - 4 \int_{3}^{3 / 2} \frac{{(p - 1)}^{2} d p}{4 p^{2}} = - \int_{3}^{3 / 2} \frac{p^{2} + 1 - 2 p}{p^{2}} d p$

$= - \int_{3}^{3 / 2} (1 + \frac{1}{p^{2}} - \frac{2}{p}) d p = - {[p - \frac{1}{p} - 2 \log p]}_{3}^{3 / 2}$

$= - [{\frac{5}{6} - 2 \log_{e} (\frac{3}{2})} - {\frac{8}{3} - 2 \log_{e} 3}]$

$= - [2 \log_{e} 2 - \frac{11}{6}] = \frac{11}{6} - \log_{e} 2^{2} = \frac{11}{6} - \log_{e} 4$

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