∫1/(x²+y²)∧(3/2))dx怎么算?

∫1/(x²+y²)∧(3/2))dx怎么算?
∫1/(x²+y²)∧(3/2))dx怎么算?

∫1/(x²+y²)∧(3/2))dx怎么算?
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/y²)x/√(x²+y²) + C
=x/[y²√(x²+y²)] + C
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