Question

How do you integrate `(x-1) e^(-x^2+2x) dx`?

Answer 1

`- 1/2 e^(-x^2+2x) + C`

Explanation:

`int color(red)((x-1) e^(-x^2+2x) )\ dx`

the easy way, look for the pattern

`d/dx ( e^(-x^2+2x) )`

`=d/dx(-x^2+2x) * e^(-x^2+2x)`

`=(-2x+2) * e^(-x^2+2x)`

`=-2 color(red)((x-1) e^(-x^2+2x))`

so
`int (x-1) e^(-x^2+2x) \ dx`

`=int - 1/2 d/dx ( e^(-x^2+2x) ) \ dx`

`= - 1/2 e^(-x^2+2x) + C`

Answer 2

`-1/2e^(_(x^2-2x)) +C`

Explanation:

Let, `I=int(x-1)e^(-x^2+2x)dx`

We take subst. `-x^2+2x=t, so, (-2x+2)dx=-2(x-1)dx=dt`, or,

`(x-1)dx=-1/2dt`.

Therefore,

`I=int(x-1)e^(-x^2+2x)dx=-1/2inte^tdt=-1/2e^t`

`=-1/2e^(-x^2+2x)=-1/2e^(_(x^2-2x)) +C`