求不定积分∫ x^2*e(x^2) dx？
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∫x^2e(x^2)dx=(1/2)∫xe^(x^2)dx^2=(1/2)∫xd(e^x^2)=(1/2)xe^(x^2)-(1/2)∫e^x^2d(x^2)^(1/2)
=(1/2)xe^(x^2)-(1/4)∫e^x^2dx^2/(x^2)^(1/2)
∫e^x^2dx^2/(x^2)^(1/2)
取t=(x^2)
=∫e^tdt/t^(1/2)
e^t=1+t+t^2/2!+t^3/3!+..+t^n/n!
=∫[1/t^(1/2)+t^(1/2)+t^(3/2)/2!+t^(5/2)/3!+..+t^(n-1/2)/n!]dt
=2t^(1/2)+(2/3)t^(3/2)+(2/5)t^(5/2)/2!+(2/7)t^(7/2)/3!+..+(n+1/2)*t^(n+1/2)/n!+C
∫x^2e^(x^2)dx
=(1/2)xe^(x^2)-(1/4)[2*x+(2/3)x^3+(2/5)x^5/2!+(2/7)x^7/3!+..+(n+1/2)x^(2n+1)/n!] +C,2,
=(1/2)xe^(x^2)-(1/4)∫e^x^2dx^2/(x^2)^(1/2)
∫e^x^2dx^2/(x^2)^(1/2)
取t=(x^2)
=∫e^tdt/t^(1/2)
e^t=1+t+t^2/2!+t^3/3!+..+t^n/n!
=∫[1/t^(1/2)+t^(1/2)+t^(3/2)/2!+t^(5/2)/3!+..+t^(n-1/2)/n!]dt
=2t^(1/2)+(2/3)t^(3/2)+(2/5)t^(5/2)/2!+(2/7)t^(7/2)/3!+..+(n+1/2)*t^(n+1/2)/n!+C
∫x^2e^(x^2)dx
=(1/2)xe^(x^2)-(1/4)[2*x+(2/3)x^3+(2/5)x^5/2!+(2/7)x^7/3!+..+(n+1/2)x^(2n+1)/n!] +C,2,
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