Prove that `int_0^x e^(x t-t^2)` `dt=e^(x^(2)/4)int_0^x e^-(t^(2)/4)dt`

Explanation

Let `I=int_(0)^(x)e^(xt)e^(-t^(2))dt`
`=e^(x^(2)//4) int_(0)^(x)e^(-x^(2)//4)e^(xt)e^(-t^(2))dt`
`=e^(x^(2)//4)int_(0)^(x)e^(-(x^(2)//4-tx+t^(2)))dt`
`=e^(x^(2)//4)int_(0)^(x)e^(-x//2-t^(2))dt`
The result clearly suggests that we have to substitute `y//2` for `x//2-t`.
Then `dt=-dy//2`. Also when `t=0,y=x`, and when `t=x, y=-x`. Thus,
`I=e^(x^(2)/4)int_(x)^(-x)e^(-y^(2)//4)(-dy//2)`
`=(e^(x^(2)//4))/2int_(-x)^(x)e^(-y^(2)//4)dy`
`(e^(x^(2)//4))/2 2int_(0)^(x)e^(-y^(2)//4)dy` [`e^(y^(2)//4)` is an even function]
`=e^(x^(2)//4)int_(0)^(x)e^(-t^(2)//4)dt`