求不定积分∫dx/[x^2√(x^2-1)]和∫dx/[x√(1-x^2)]推导详细些,顺便说一下解法,碰到这种类型的怎么解.

Question

求不定积分∫dx/[x^2√(x^2-1)]和∫dx/[x√(1-x^2)]
推导详细些,顺便说一下解法,碰到这种类型的怎么解.

Answer 1

∫1/[x√(x^2-1)]dx
=∫(1/x^2)/[√(x^2-1)/x]dx
=∫(1/x^2)dx/√[1-(1/x)^2]
= -∫d(1/x)/√[1-(1/x)^2]
= -arcsin(1/x)+C
其中C为任意常数
∫ 1/[x√(1-x²)] dx
分子分母同乘以x
=∫ x/[x²√(1-x²)] dx
=(1/2)∫ 1/[x²√(1-x²)] d(x²)
令√(1-x²)=u,则x²=1-u²,d(x²)=-2udu
=(1/2)∫ 1/[(1-u²)u](-2u)du
=∫ 1/(u²-1) du
=(1/2)ln|(1-u)/(1+u)| + C
=(1/2)ln|(1-√(1-x²))/(1+√(1-x²))| + C
=(1/2)ln|(1-√(1-x²))²/x²| + C
=ln|(1-√(1-x²))/x| + C