求不定积分S（1-X）/√(9-x^2)dx, S e^√x dx , S x^2 lnx dx , S x cos2x dx , S x e^-x dx

Question

别直接写结果啊

Answer 1

∫(1-x)/√(9-x²) dx
= ∫dx/√(3²-x²) - ∫xdx/√(9-x²)
= arcsin(x/3) - (1/2)∫d(x²)/√(9-x²)
= arcsin(x/3) + (1/2)∫d(9-x²)/√(9-x²)
= arcsin(x/3) + √(9-x²) + C

∫e^√x dx，u²=x，2udu=dx
= 2∫u*e^u du = 2∫u de^u
= 2u*e^u - 2∫e^u du
= 2u*e^u - 2e^u + C
= (2u-1)*e^u + C
= (2√x-1)*e^√x + C

∫x²lnx dx
= ∫lnx d(x³/3)
= (x³lnx)/3 - (1/3)∫x³ dlnx
= (x³lnx)/3 - (1/3)∫x³(1/x) dx
= (x³lnx)/3 - (1/3)(x³/3) + C
= (x³/9)(3lnx - 1) + C

∫xcos2x dx
= (1/2)∫xcos2x d(2x) = (1/2)∫x dcos2x
= (xcos2x)/2 - (1/2)∫cos2x dx
= (xcos2x)/2 - (1/4)∫cos2x d(2x)
= (xcos2x)/2 - (1/4)sin2x + C
= (1/4)(2xcos2x - sin2x) + C

∫xe^-x dx
= -∫xe^-x d(-x) = -∫x de^-x
= -xe^-x - [-∫e^-x dx]
= -xe^-x - [∫e^-x d(-x)]
= -xe^-x - [e^-x] + C
= -(x+1)*e^-x + C

Answer 2

1 )arcsin(x/3)+√(9-x^2)+c
2)2√xe^√x-2e^√x+c
3)1/3Inx*x^3-1/9x^3+c
4)1/2x*sin2x+1/4cos2x+c
5)-xe^-x-e^-x+c追问

过程呢？

追答

兄弟，过程不好写
1）换元3cosa=√(9-x^2),3sina=x
2)3)4)5)均是分部积分，对x进行将次

Answer 3

1.令x=3cost;∫ (1-x)/√(9-x²)dx=∫(1-3cost)/3sintd(3cost)=∫-1+3costdt=3sint-t=3√(9-x²)-arccos(x/3)+c
2.令t=√x; ∫ e^√xdx=∫ e^tdt²=∫ 2t*e^tdt=2t*e^t-2e^t+c=2√x*e^√x-2e^√x+c
3.∫x²lnxdx=(1/3)∫lnxdx³=(1/3)x³lnx-(1/3) ∫ x³dlnx+c=(1/3)x³lnx-(1/3) ∫ x²dx+c=(1/3)x³lnx-(1/9) x³+c
4.∫xcos2xdx=(xsin2x)/2+(cos2x)/4+c
5.∫xe^(-x)dx=-x/(e^x)-1/(e^x)+c

Answer 4

1 )arcsin(x/3)+√(9-x^2)+c
2)2√xe^√x-2e^√x+c
3)1/3Inx*x^3-1/9x^3+c
4)1/2x*sin2x+1/4cos2x+c
5)-xe^-x-e^-x+c