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∫E^x^1/3Dx

令u=x^(1/3) 则 x=u^3,dx=3u^2du 于是∫e^[x^(1/3)]dx=3∫u^2*e^udu=3∫u^2de^u =3u^2*e^u-6∫ue^udu=3u^2*e^u-6∫ude^u =3u^2*e^u-6ue^u+6∫e^udu=3(u^2-2u+2)e^u+C =3[x^(2/3)-2x^(1/3)+2]e^[x^(1/3)]+C 设G(x)是f(x)的另一个原函数,即∀x∈I...

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如图所示:

∫(√3->√5) dx/√(e^x -1) let e^(x/2) = secu (1/2)e^(x/2) dx = secu.tanu du dx = 2tanu du x=√3, u=arctan√(e^(√3) -1) x=√5, u=arctan√(e^(√5) -1) ∫(√3->√5) dx/√(e^x -1) =∫(arctan√(e^(√3) -1)->arctan√(e^(√5) -1)) 2du =2[ arctan√(e^(√...

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