Question

(x^2 1)/(x^4 1)的不定积分

(x^2+1)/(x^4+1)不知道为什么提交后加号没了

Answer 1

∫1/（x^4-1）dx
=1/2∫[1/（x^2-1）-1/（x^2+1）]dx
=1/2∫[1/2*1/（x-1）-1/2*1/(x+1)-1/（x^2+1）]dx
1/4ln(x-1)-1/4ln(x+1)-1/2arctanx+C

Answer 2

换元脱根号
令u=√(1－x)
=∫1/(1+u²)ud(1－u²)
=－2∫1/(1+u²)du
=－2arctanu+C

Answer 3

∫xdx/(1+x^4)^2 = (1/2)∫d(x^2)/(1+x^4)^2 (令 x^2 = tanu)
= (1/2)∫(secu)^2du/(secu)^4 = (1/2)∫(cosu)^2du
= (1/4)∫(1+cos2u)du = (1/4)[u+(1/2)sin2u] + C
= (1/4) [arctan(x^2) + x^2/√(1+x^4)] + C

Answer 4

∫[1/(4-x2)]dx =?∫[1/(x+2) -1/(x-2)]dx =?[ln|x+2|-ln|x-2|] +C =?ln|(x+2)/(x-2)| +C

Answer 5

∫[(x+1)/(x²+1)]dx
=∫[x/(x²+1) +1/(x²+1)]dx
=(1/2)ln(x²+1) +arctanx +C