Mathematics

Integrate with respect to $$\dfrac { \sqrt { { x }^{ 2 }-8 }  }{ { x }^{ 4 } } $$


SOLUTION

Consider the given integral.

$$ I=\int{\dfrac{x-8}{{{x}^{4}}}dx} $$

$$ I=\int{\left( \dfrac{1}{{{x}^{3}}}-\dfrac{8}{{{x}^{4}}} \right)dx} $$

$$ I=\int{\dfrac{1}{{{x}^{3}}}dx}-8\int{\dfrac{1}{{{x}^{4}}}dx} $$

$$ I=-\dfrac{1}{2{{x}^{2}}}+\dfrac{8}{3{{x}^{3}}}+C $$

$$ I=\dfrac{16-3x}{6{{x}^{3}}}+C $$

 

Hence, this is the required value of the integral.
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