Question

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Answer 1

$\int \frac{ {x}^{2} }{16 - {x}^{6} } \: dx$

$= \int \frac{ {x}^{2} }{16 - { ( {x}^{3}) }^{2} } \: dx$

$t = {x}^{3} = > dt = ( {x}^{3}) ' = 3 {x}^{2} \: dx$

$= \frac{1}{3} \int \frac{ 3{x}^{2} }{16 - {t}^{2} } = \frac{1}{3} \int \frac{dt}{16 - {t}^{2} }$

$= - \frac{1}{3} \int \frac{dt}{ {t}^{2} - 16} = - \frac{1}{3} \times \frac{1}{2 \times 4} \times ln | \frac{t - 4}{t + 4} |$

$= - \frac{1}{24} \times ln | \frac{ {x}^{3} - 4}{ {x}^{3} + 4} |$

$= - \frac{ln | {x}^{3} - 4| - ln | {x}^{3} + 4| }{24} = \frac{ln | {x}^{3} + 4| - ln | {x}^{3} - 4 | }{24} + C$