∫1/(x²+y²)∧(3/2))dx怎么算?
问题描述：
∫1/(x²+y²)∧(3/2))dx怎么算?
答
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答
y看做常数令x=ytanu,则：(x²+y²)^(3/2)=y³sec³u,dx=ysec²udu∫1/(x²+y²)^(3/2))dx=∫ ysec²u/(y³sec³u) du=(1/y²)∫ cosu du=(1/y²)sinu + C=(1/y...