Question

求∫(5-x^2)^(-1/2)dx不定积分,麻烦写下过程，谢谢

Answer 1

利裤判用反正弦函数的求导胡嫌改公式，可知者晌

Answer 2

∫基手(5-x^2)^(-1/2)dx = ∫谈锋宏dx/√(5-x^2) = (1/√5)∫dx/√[1-(x/√5)^2]
= ∫d(x/含册√5)/√[1-(x/√5)^2] = arcsin(x/√5) + C