求x^2/(a^2-x^2)^1/2dx的不定积分

必须用变量换元法

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不定积分x^2/(a^2-x^2)^1/2dx
=-积分[a^2-x^2-a^2]/(a^2-x^2)^1/2dx
=-积分(a^2-x^2)^1/2dx+积分a^2/绝携(a^2-x^2)^1/芦宏宏2dx
=-x(a^2-x^2)^1/2-积分x^2/(a^2-x^2)^1/陪册2dx+a^2arcsin(x/a)+C
得不定积分x^2/(a^2-x^2)^1/2dx
=-(x/2)(a^2-x^2)^1/2+(a^2/2)arcsin(x/a)+C

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设x=sinu
结果：a^2[u-(1/2)sin2u]/2+C

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