∫2r^4e^-r^2dr的原导是多少

原函数不能用初等函数表示.
令 u=r^2,
I =∫2r^4e^(-r^2)dr =∫r^3e^(-r^2)dr^2 = -∫r^3de^(-r^2) = -∫u^(3/2)de^(-u)
= -u^(3/2)e^(-u) +(3/2)∫u^(1/2)e^(-u) du
=- u^(3/2)e^(-u) -(3/2)∫u^(1/2)de^(-u)
= -u^(3/2)e^(-u) -(3/2)u^(1/2)e^(-u) +(3/2)∫[e^(-u)/(2√u)]du
= -x(1+x^2)e^(-x^2)+(3/2)∫e^(-x^2)dx
∫e^(-x^2)dx 不能用初等函数表示!

求不定积分∫2r⁴e^(-r²)dr
原式=-∫r³d[e^(-r²)]=-[r³e^(-r²)-3∫r²e^(-r²)dr]
=-r³e^(-r²)-(3/2)∫rde^(-r²)
=-r³e^(-r²)-(3/2)[re^(-r²...