Question

1/(1-x^2)^3/2的不定积分是怎么求的啊，急，还有x^3/(1+x^2)3/2的不定积分

Answer 1

∫dx/√(1-x^2)^3
=∫dx/[x^3√(1/x^2-1)^3
=(-1/2)∫d(1/x^2)/√(1/x^2-1)^3
=1/√(1/x^2-1)+C

∫x^3dx/(1+x^2)^(3/2)
=(1/2)∫x^2d(x^2)/(1+x^2)^(3/2)
=(1/2)∫dx^2/(1+x^2)^(1/2)-(1/2)∫dx^2/(1+x^2)^(3/2)
=√(1+x^2)+1/√(1+x^2)+C

追问

可以用第二换元法解吗，这个不太明白

追答

1
x=sinu
∫du/cosu^2=tanu=x/√(1-x^2)=1/√(1/x^2-1)

2
x=tanu
=∫tanu^3du/secu
=∫sinu^3du/cosu^2
= -∫sinu^2dcosu/cosu^2
=∫(cosu^2-1)dcosu/cosu^2
=cosu+1/cosu
=1/√(1+x^2)+√(1+x^2)