求不定积分的问题谢谢

(1)
let
u=e^x
du= e^x dx
∫ xe^x/(1+e^x)^2 dx
=∫ lnu /(1+u)^2 du
=-∫ lnu d[1/(1+u)]
=-lnu/(1+u) + ∫ du/[u(1+u)]
=-lnu/(1+u) + ∫ [1/u-1/(1+u)] du
=-lnu/(1+u) + ln|u| - ln|1+u| + C
=-x/(1+e^x) + x - ln|1+e^x| + C
(2)
f(x) =x(cosx)^2
f(-x)= -f(x)
∫(-π/2->π/2) (x+cosx)^2.cosx dx
=∫(-π/2->π/2) [(x^2+(cosx)^2 ].cosx dx +2∫(-π/2->π/2) x(cosx)^2 dx
=∫(-π/2->π/2) [(x^2+(cosx)^2 ].cosx dx + 0
=2∫(0->π/2) [(x^2+(cosx)^2 ].cosx dx
=2∫(0->π/2) x^2. cosx dx +2∫(0->π/2) (cosx)^3 dx
=2∫(0->π/2) x^2 dsinx +2∫(0->π/2) [ 1-(sinx)^2] dsinx
=2[ x^2.sinx]|(0->π/2) -4∫(0->π/2) xsinx dx + 2[ sinx -(1/3)(sinx)^3]|(0->π/2)
=(1/2)π^2 +4∫(0->π/2) xdcosx + 4/3
=(1/2)π^2 + 4/3 + 4[xcosx]|(0->π/2) - 4∫(0->π/2) cosx dx
=(1/2)π^2 + 4/3 - 4[sinx]|(0->π/2)
=(1/2)π^2 + 4/3 + 4
=(1/2)π^2 + 16/3