.∫[x^3/(3+x)]dx
=∫ [(x^2-3)+ 9/(x+3) ] dx
= x^3/3 -3x + 9ln|x+3| + C
.∫dx/(1+cosx)
=(1/2)∫ 1/[cos(x/2)]^2 dx
= (1/2)∫ [sec(x/2)]^2 dx
= tan(x/2) + C第一題=∫ [(x^2-3)+ 9/(x+3) ] dx 通分回去和原式不一樣啊第二題最后那個(gè)1/2為什么不要謝謝(1).∫[x^3/(3+x)]dx =.∫[ x^2-3x+9-27/(3+x)]dx= x^3/3 -3x^2+9x -27ln|3+x| +C(2).∫dx/(1+cosx)=(1/2)∫1/[cos(x/2)]^2dx= (1/2)∫ [sec(x/2)]^2 dx= ∫ [sec(x/2)]^2 d(x/2)= tan(x/2) + C