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∫E∧ 3x 9 Dx

令t=√(3x+9),则x=t/3-3,dx=2/3t,所以∫e^√(3x+9)dx =2/3∫te^tdt=2/3(t-1)e^t+C=2/3(√(3x+9)-1)e^√(3x+9)+C

令t=√(3x+9),则x=t/3-3,dx=2/3t,所以∫e^√(3x+9)dx =2/3∫te^tdt=2/3(t-1)e^t+C=2/3(√(3x+9)-1)e^√(3x+9)+C

令√(3x+9)=u,则:3x+9=u^2,∴x=(1/3)u^2-3,∴dx=(2/3)udu.∴∫{e^[√(3x+9)]}dx =(2/3)∫ue^udu =(2/3)∫ud(e^u) =(2/3)ue^u-(2/3)∫e^udu =(2/3)√(3x+9)e^[√(3x+9)]-(2/3)e^u+C =(2/3)√(3x+9)e^[√(3x+9)]-(2/3)e^[√(3x+9)]+C.

设根号(3x+9)=t,则dx=(2t/3)dt∫e^(根号下(3x+9))dx=∫(e^t)(2t/3)dt=(2/3)∫(e^t)tdt=(2/3)∫td(e^t)=(2/3)(te^t-e^t)+C=(2/3)(t-1)e^t+C

∫e^3xdx=1/3e^3xd3x 令t=3x 原式=e^tdt=1/3e^t+C=1/3e^3x+C

-1/3*e^(-3x)+d

1. 若是 ∫[e^(3x^2)+lnx]dx = ∫e^(3x^2)dx + ∫lnxdx 前者不能用初等函数表示,故本题积不出来. 若是 ∫e^(3x^2+lnx)dx = ∫xe^(3x^2)dx = (1/6)∫e^(3x^2)d(3x^2) = (1/6)e^(3x^2)+C.2. ∫x(x-1)^9dx = (1/10)∫xd(x-1)^10 = x(x-1)^10/10-(1/10)∫(x-1)^10dx = x(x-1)^10/10 - (x-1)^11/110 + C

不停地分部积分.∫e^(3x)sinxdx=-∫e^(3x)d(cosx)=-e^(3x)cosx+∫cosxd(e^(3x))=-e^(3x)cosx+3∫e^(3x)cosxdx=-e^(3x)cosx+3∫e^(3x)d(sinx)=-e^(3x)cosx+3e^(3x)sinx-3∫sinxd(e^(3x))=-e^(3x)cosx+3e^(3x)sinx-9∫e^(3x)sinxdx注意到右边最后一项和左边是一样的,就可以解出:∫e^(3x)sinxdx=e^(3x)(3sinx-cosx)/10+C

令√(3x+9)=u,则:3x+9=u^2,∴3dx=2udu,∴dx=(2/3)udu.∴∫[e^√(3x+9)]dx=(2/3)∫ue^udu=(2/3)∫ud(e^u)=(2/3)ue^u-(2/3)∫(e^u)du=(2/3)√(3x+9)e^√(3x+9)-(2/3)e^u+C=(2/3)√(3x+9)e^√(3x+9)-(2/3)e^√(3x+9)+C

答:换元令x^=t,则x=t^2 ∫e^x^ dx=∫e^td(t^2)=∫2te^tdt=2te^t-2e^t+c 将t=x^代入得:∫e^x^ dx=2x^*e^x^ - 2e^x^ +c

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