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

caicaizi_20072022-10-04 11:39:544条回答

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
别直接写结果啊

已提交,审核后显示!提交回复

共4条回复
七月javesky 共回答了15个问题 | 采纳率100%
∫(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
1年前
ahhf315 共回答了7个问题 | 采纳率
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
1年前
963138 共回答了18个问题 | 采纳率
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.∫x²l...
1年前
new25 共回答了56个问题 | 采纳率
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
1年前

相关推荐

大家在问