Question

求大佬用凑微分法，三角函数代换法，倒代法，简单无理函数代换法，解∫(1/x√(4-x²）)dx

Answer 1

∫dx/[x√(4-x^2)]
let
x=2sinu
dx=2cosu du
∫dx/[x√(4-x^2)]
=∫du/(2sinu)
=(1/2)∫cscu du
=(1/2)ln| cscu - cotu| +C
=(1/2)ln|2/x - √(4-x^2)/x| +C