不定积分x四次方分之根号下x²-a²11

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摘要令 $x = a\sec\sqrt{x^{2} - a^{2}} = a\tan t$

则 $dx = a\sec t \tan t dt$

$\int\sqrt{x^{2} - a^{2}}/x^{4}dx = \int\left[a^{2}\sec t \left(\tan t\right)^{2}/a^{4}\sec^{4}x\right]dt$

$= \frac{1}{a^{2}}\int\left(\tan t\right)^{2}\left(\cos t\right)^{3}dt$

$= \frac{1}{a^{2}}\int\left[\sec t - 1\right]\left(\cos t\right)^{3}dt$

$= \frac{1}{a^{2}}\left[\int\cos t dt - \int\left(\cos t\right)^{3}dt\right]$

$= \frac{1}{a^{2}}\left[\int\cos t dt - \int\left(1 - \sin t^{2}\right)d\sin t\right]$

$= \frac{1}{3a^{2}}\left(\sin t\right)^{3} + c$

$= \frac{1}{3a^{2}}\left(\sqrt{x^{2} - a^{2}}/x\right)^{3} + c$

咨询记录 · 回答于2023-12-22

不定积分x四次方分之根号下x²-a²11

或者写在纸上发给我哦

（10）

令 $x = a \sec \sqrt{x^2 - a^2}$ 令 $dx = a \sec t \tan t dt$ $\int \frac{\sqrt{x^2 - a^2}}{x^4} dx = \int \left[ \frac{a^2 \sec t (\tan t)^2}{a^4 (\sec t)^4} \right] dt$ $= \frac{1}{a^2} \int (\tan t)^2 (\cos t)^3 dt$ $= \frac{1}{a^2} \int \left[ (\sec t)^2 - 1 \right] (\cos t)^3 dt$ $= \frac{1}{a^2} \left[ \int (\cos t) dt - \int (\cos t)^3 dt \right]$ $= \frac{1}{a^2} \left[ \int (\cos t) dt - \int (1 - (\sin t)^2) d(\sin t) \right]$ $= \frac{1}{3a^2} (\sin t)^3 + c$ $= \frac{1}{3a^2} \left( \sqrt{x^2 - a^2} / x \right)^3 + c$

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