求x^2√(x^2-a^2)的不定积分 求1/(x^2+a^2)的不定积分
\u6c42\u4e0d\u5b9a\u79ef\u5206 x^2/\u221a(a^2-x^2) dx \u6c42\u8fc7\u7a0b \u8c22\u8c22R=\u539f\u5f0f=\u222b[(x²-a²+a²)/\u221a(a^2-x^2)]dx
=-\u222b\u221a(a^2-x^2)dx+a²\u222b[1/\u221a(a^2-x^2)]dx=P+Q
\u524d\u4e00\u9879\u79ef\u5206P\u53ef\u4ee5\u5229\u7528\u5206\u6b65\u79ef\u5206\u6cd5\uff1a
P=[-x\u221a(a^2-x^2)]+\u222bxd[\u221a(a^2-x^2)]=[-x\u221a(a^2-x^2)]-\u222b[x²/\u221a(a^2-x^2)]dx=[-x\u221a(a^2-x^2)]-R
\u540e\u4e00\u9879\u79ef\u5206Q\u4f7f\u7528\u7b2c\u4e00\u7c7b\u6362\u5143\u79ef\u5206\u6cd5\u53ef\u4ee5\u6c42\u5f97\uff1a
Q=a²arcsin(x/a)+C0.......................................(C0\u4e3a\u5e38\u6570)
\u7efc\u4e0a\u6709\uff1aR=P+Q=[-x\u221a(a^2-x^2)]-R+a²arcsin(x/a)+C0
\u89e3\u5f97\uff1a\u539f\u5f0f=R=(1/2){[-x\u221a(a^2-x^2)]+a²arcsin(x/a)}+C...........................(C=0.5C0)
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