How do you integrate $(x - 1) e^{- x^{2} + 2 x} d x$ $(x-1) e^(-x^2+2x) dx$ ?

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Answer 1 · 2016-07-31T15:11:16

$- \frac{1}{2} e^{- x^{2} + 2 x} + C$ $- 1/2 e^(-x^2+2x) + C$

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

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 · 2016-07-31T15:19:16

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

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,

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

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

How do you integrate (x-1) e^(-x^2+2x) dx(x−1)e−x2+2xdx (x-1) e^(-x^2+2x) dx?