Integrating xe^-(x-2): Solve with Substitution

[ramses07]

I'm having a brain fart and I can't figure out

xe^-(x-2).

I tried integrating by sub., which led me to;

U = x^2-4x+4
du = 2x -4

But that doesn't solve it, can anybody tell me ?

 

[Gregg]

You don't need to use substitution really. Write the exponential as a product of two terms.

 

[magicarpet512]

\int xe^{-(x-2)}dx

I tried integrating by sub., which led me to;

U = x^2-4x+4
du = 2x -4

Your "U" doesn't really make sense.
Did you think to try integration by parts?

 

[HallsofIvy]

Is it possible that you meant
\int xe^{-(x-2)^2}dx

If that is the case, then, yes, the substitution u= (x- 2)^2= x^2- 4x+ 4 is reasonable. du= (2x- 4) dx which you can use by rewriting the integral as
\frac{1}{2}\int [(2x- 4)e^{-(x-2)^2}+ 4e^{-(x-2)^2}]dx

\frac{1}{2}\int e^{-u}du+ 2\int e^{-(x-2)^2}dx
Now that first integral is easy but the second integral can not be done in terms of elementary functions.