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i really dont have an idea...
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Other answers:

Neither do I.. :( Let u=x+4 du = dx => x+2 = u-2 ; x = u-4 \[\int (\frac{x+2}{x+4})^2 e^x dx = \int (\frac{u-2}{u})^2 e^{u-4}du = \int\frac{u^2-4u+4}{u^2}e^{u-4}du\]\[=\int (e^{u-4}-\frac{4e^{u-4}}{u}+\frac{4e^{u-4}}{u^2})du\] I'm sure @hartnn has a better idea!!
can u find derivative of x/(x+4) ??
yes...wait...but why x / x+4?
find its derivative, u'll come to know.....
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wait i think i made a mistake....
u'll need integration by parts. but that becomes very messy...
i just realized that x/x+4 + d/dx (x/x+4) = [(x+2)/(x+4)]^2
then using this : \(\\ \huge \color{red}{\int e^x[f(x)+f’(x)]dx=e^xf(x)+c} \\\) i got the integral as (xe^x)/(x+4) + c
i don't know how to explain that ....
Reverse: \[\frac{d}{dx} e^x f(x)+C= e^xf(x)+e^xf'(x)=e^x(f(x)+f'(x))\]So, \[\int e^x(f(x)+f'(x)) = e^xf(x)+C\] And the question, Consider f(x) = x/x+4 f'(x) = 4/(x+4) f(x) + f'(x) = [(x+2)/(x+4)]^2 So, \[\int [\frac{(x+2)}{(x+4)}]^2e^xdx = \int e^x(\frac{x}{x+4}+\frac{4}{(x+4)^2})dx=e^x(\frac{x}{x+4})+C\] Hail @hartnn
yes..that's im about to type lol thanks! @hartnn @RolyPoly
i meant i can't explain how we can think of f(x) as x/(x+4) all other steps are just use of formula.....
comes with practice, i guess...

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