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

2 Answers
Jul 31, 2016

Answer:

# - 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#

Jul 31, 2016

Answer:

#-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#