这个不定积分怎么算啊

如下

如下图片

(5)

∫ dx/[ 1+ (x+2)^(1/3) ]
let
(x+2)^(1/6) = tanu
(1/6)(x+2)^(-5/6) dx = (secu)^2 du
dx = 6(tanu)^5.(secu)^2 du
∫ dx/[ 1+ (x+2)^(1/3) ]
=∫ 6(tanu)^5.(secu)^2 du/ (secu)^2
=6∫ (tanu)^5 du
=6∫ (tanu)^3 .[(secu)^2 -1 ] du
=6∫ (tanu)^3 dtanu - 6∫ (tanu)^3 du
=(3/2)(tanu)^4 - 6∫ (tanu).[ (secu)^2 -1] du
=(3/2)(tanu)^4 - 6∫ (tanu) dtanu +6∫ tanu du
=(3/2)(tanu)^4 - 3(tanu)^2 -6ln|cosu| + C
=(3/2)(x+2)^(2/3) - 3(x+2)^(1/3) -6ln| 1/√[1+ (x+2)^(1/3)]| + C
=(3/2)(x+2)^(2/3) - 3(x+2)^(1/3) +3ln|1+ (x+2)^(1/3)| + C
(6)
x= tanu
dx= (secu)^2 du
∫ arctanx/(1+x^2)^(3/2) dx

= ∫ [u / (secu)^3 ] [(secu)^2 du]
=∫ u cosu du
=∫ u dsinu
=usinu -∫ sinu du
=usinu + cosu + C
= (arctanx)[x/√(1+x^2)] + 1/√(1+x^2) + C
=[xarctanx +1]/√(1+x^2) + C

小猿搜题作业帮更快 还能有免费视频讲解