Question

求不定积分∫dx/[x^2√(x^2-1)]和∫dx/[x√(1-x^2)]

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