Question

求不定积分∫dx/√1+x2.x^4

Answer 1

设x=tanθ，则dⅹ=sec²θdθ.
同时，sinθ=ⅹ/√(1+x²).
∴∫[1/√(1+x²)]dx
=∫(1/secθ)sec²θdθ
=∫secθdθ
=∫(1/cos²θ)d(sinθ)
=1/2∫[1/(1+sinθ)+1/(1-sinθ)]d(sinθ)
=1/2㏑(1+sinθ)-1/2㏑(1-sin)+C
=1/2㏑[(1+sinθ)/(1-sinθ)]+C
=1/2㏑[(x+√(1+x²))/(-x+√(1+x²))]+C
=㏑[x+√(1+x²)]+C。

Answer 2

换元x=tanu
=∫secu/(tanu)^4du
=∫cos²u/(sinu)^4dsinu
=∫1/(sinu)^4-1/sin²udsinu
=1/sinu-1/3sin³u+C

追问

结果呢

追答

x=tanu，sinu=tanu/secu=x/√(1-x²)代回即可。