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116问答网 > 你好，打扰一下，请问（a^2+x^2）^-1⼀2的原函数是多少？

你好，打扰一下，请问（a^2+x^2）^-1⼀2的原函数是多少？

2025-03-16 17:18:36

推荐回答（1个）

回答1：

a≠0
令x=atant→tant=x/a→sect=√1+(x/a)²=√(a²+x²)/a
dx=asec²tdt
a²+x²=a²(1+tan²t)=(asect)²
∫dx/√(1+x²)
=∫(a·sec²t/asect·dt
=∫sectdt
=ln|sect＋tant|＋C
=ln|√(a²+x²)+x|/|a|+C
a=0
∫dx/√x²=ln|x|+C

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