How do you integrate $(e^(x-2))^2/e^(2x)$ ?

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Answer 1 · 2016-10-01T02:50:41

Use properties of exponents to rewrite.

$(e^(x-2))^2 = e^(2(x-2)) = e^(2x-4) = e^(2x)e^(-4)$

So $(e^(x-2))^2/e^(2x) = (e^(2x)e^-4)/e^(2x) = e^-4$

Thus $int (e^(x-2))^2/e^(2x) dx = int e^-4 dx = x e^-4 +C$

Answer 2 · 2016-10-01T02:59:48

$x/e^4 + c$

$int(e^(x-2))^2/e^(2x) dx = int(e^(2x-4))/e^(2x) dx$

$= inte^(2x)/(e^(2x)*e^4) dx$

$= int cancel(e^(2x))/(cancel(e^(2x))e^4) dx$

Thus the integral reduces to:

$int 1/e^4 dx$

Since $1/e^4$ is a constant:

$= 1/e^4 int dx = 1/e^4*x +c$

$=x/e^4 + c$

How do you integrate (e^(x-2))^2/e^(2x)(e^(x-2))^2/e^(2x)?