作业帮求不定积分1.∫x^3/(3+x)dx 2.∫dx/(1+cosx)
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/16 17:48:41
求不定积分1.∫x^3/(3+x)dx 2.∫dx/(1+cosx)
求不定积分1.∫x^3/(3+x)dx 2.∫dx/(1+cosx)
求不定积分1.∫x^3/(3+x)dx 2.∫dx/(1+cosx)
.∫[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
1.∫[x^3/(3+x)]dx
=∫ [(x^2-3x+9)dx-∫27/(x+3) ] dx
= x^3/3 -3x^2/2 + 9x-27ln|x+3| + C
2. ∫dx/(1+cosx)
= 1/2∫1/[cos(x/2)]^2 dx
= 1/2∫[sec(x/2)]^2 dx
=∫[sec(x/2)]^2 d(x/2)
= tan(x/2) + C
1.∫[(x³+27-27)/(3+x)]dx
=.∫[(x³+3³)-27)/(3+x)]dx
=.∫[(x+3)(x^2-3x+9)-27)/(3+x)]dx
=.∫ {[(x^2-3x+9)]-27)/(3+x)}dx
=.∫ {[(x^2-3x+9)]dx-27∫1/(3+x)}dx
=.∫ {[(1/3)x³...
全部展开
1.∫[(x³+27-27)/(3+x)]dx
=.∫[(x³+3³)-27)/(3+x)]dx
=.∫[(x+3)(x^2-3x+9)-27)/(3+x)]dx
=.∫ {[(x^2-3x+9)]-27)/(3+x)}dx
=.∫ {[(x^2-3x+9)]dx-27∫1/(3+x)}dx
=.∫ {[(1/3)x³-(3/2)x^2+9x)]'dx-27∫[lnl(3+xl]'dx
=(1/3)x³-(3/2)x^2+9x+C1-27lnl3+xl+C2
=(1/3)x³-(3/2)x^2+9x-27lnl3+xl+C
2.∫dx/(1+cosx)
=∫(1/2)[sec(x/2)]^2dx
=∫[sec(x/2)]^2d(x/2)
=∫dtan(x/2)
=tan(x/2)+c
收起