Question

怎么求e^(x^2-x)的不定积分

Answer 1

e^(-x^2)积分：
1.如果积分限是－∞到∞,∫e^(-x^2)dx =√π
2.如果积分限0到∞,根据偶函数的性质可知,∫e^(-x^2)dx =√π/2

Answer 2

∫e^(-x^2)dx=(-1/2)∫de^(-x^2)/x=(-1/2)e^(-x^2)/x -(1/2)∫e^(-x^2)dx/x^2
=(-1/2)e^(-x^2)/x-(1/4)e^(-x^2)/x^3+(1/4)∫e^(-x^2)d(1/x^3)
=(-1/2)e^(-x^2)/x-(1/4)e^(-x^2)/x^3-(1/8)e^(-x^2)/x^4+(1/8)∫e^(-x^2)d(1/x^4)
x^2=t ∫e^(-x^2)d(1/x^4)
=∫e^(-t)d(1/t^2)=e^(-t)/t^2+∫e^(-t)dt/t^2=e^(-t)/t^2-e^(-t)/t-∫e^(-t)dt/t
e^x=1+x+x^2/2!+x^3/3!+x^4/4!+..+x^n/n!
e^(-t)=1+(-t)+(-t)^2/2!+(-t)^3/3!+..+(-t)^n/n!
∫e^(-t)dt/t=lnt-t -t^2/(2*2!)-t^3/(3*3!)-..-t^n/(n*n!)
所以
∫e^(-x^2)dx=(-1/2)e^(-x^2)/x-(1/4)e^(-x^2)/x^3-(1/8)e^(-x^2)/x^4+(1/8)e^(-x^2)/x^4-(1/8)e^(-x^2)/x^2-(1/8)[ln(x^2)-x^2-(x^2)^2/(2*2!)-(x^2)^3/(3*3!)-..-(x^2)^n/(n*n!)]

Answer 3

乘e^y^2 这个积分值和e^x^2一样得e^(x^2+y^2)然后变换成极坐标被积表达式为ΘdΘρe^ρ^2dρ= πe^ρ^2 然后开根号ρ的值和x的积分限有关

Answer 4

令x^(1/3)=t, 则dx=3t^2dt 带入积分 =∫3t^2e^tdt =∫3t^2de^t 分部积分 =3t^2e^t-∫6te^tdt =3t^2e^t-∫6tde^t =3t^2e^t-6te^t+6∫e^tdt =3t^2e^t-6te^t+6e^t+c 反带入x^(1/3)=t =3x^(2/3)e^(x^(1/3))-6x^(1/3)e^(x^(1/3))+6e^(x^(1/3))+

Answer 5

∫xe^xdx
=∫xde^x
=x*e^x-∫e^xdx
=x*e^x-e^x+C