Mathematics

$$\displaystyle \int_{2}^{4} \dfrac{\sqrt{x^{2}-4}}{x^{4}}dx=$$


ANSWER

$$\dfrac{\sqrt{3}}{32}$$


SOLUTION
$$\int_2^4 {\cfrac{{\sqrt {{x^2} - 4} }}{4}} $$
$$\int_2^4 {\cfrac{{\sqrt {1 - \cfrac{4}{{{x^2}}}dx} }}{{{x^4}}}} $$
$$\int_2^4 {\cfrac{{\sqrt {1 - \cfrac{4}{{{x^2}}}dx} }}{{{x^3}}}} $$
putting $$1 - \cfrac{4}{{{x^2}}} = t$$
$$ - 4 \times \cfrac{{ - 2}}{{{x^3}}}dx = dt$$
$$\cfrac{{dx}}{{{x^3}}} = \cfrac{{dt}}{8}$$
$$\int_0^{\cfrac{3}{4}} {\cfrac{{\sqrt t }}{8} \times dt} $$
$$ = \cfrac{1}{8} \times \cfrac{2}{3}{\left[ {{t^{\cfrac{3}{2}}}} \right]^\cfrac{3}{4}}$$
$$ = \cfrac{1}{{12}}\left[ {{{\left( {\cfrac{3}{4}} \right)}^\cfrac{3}{2}} - 0} \right]$$
$$ = \cfrac{1}{{12}}\left[ {{{\left( {\cfrac{{\sqrt 3 }}{2}} \right)}^3}} \right]$$
$$ = \cfrac{1}{{12}} \times \cfrac{{3\sqrt 3 }}{8}$$
$$ = \cfrac{{\sqrt 3 }}{{32}}$$
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Single Correct Medium Published on 17th 09, 2020
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