Question

求解一道不定积分!!!

∫e^(x^2-x)dx
感激不尽

Answer 1

不能表示为初等函数，利用刘维尔第三定理即可证明。

Answer 2

∫e^(x^2-x)dx
=(1/e^(1/4)) ∫e^(x^2-x+1/4)dx
=e^(-1/4)∫e^(x-1/2)^2d(x-1/2)
(x-1/2)=u
=e^(-1/4)∫e^u^2du
=e^(-1/4)∫de^u^2/2u
=e^(-1/4)[e^u^2/2u+∫e^u^2du/2u^2]
=e^(-1/4)[e^u^2/2u+∫de^u^2/2^2u^3]
=e^(-1/4)[e^u^2/2u+e^u^2/4u^3+∫3e^u^2du/4u^4]
=e^(-1/4)[e^u^2/2u+e^u^2/4u^3+3e^u^2/8u^5+3*5∫e^u^2du/8u^6]
=e^(-1/4)[e^u^2/2u+e^u^2/4u^3+3e^u^2/8u^5+3*5*e^u^2/16u^7+...+3*5*..*(2n-3)e^u^2/(2^n*u^(2n-1)]

=e^(-1/4)e^(x-1/2)^2[1/[2(x-1/2)]+1/[4(x-1/2)^3]+3/[8(x-1/2)^5]+...+3*5*...*(2n-3)/[2^n *(x-1/2)^(2n-1)] ] n→∞