$\int {(\frac{x + 2}{x + 4})}^{2} e^{x} d x$ is equal to

e^{x} (\frac{x}{x + 4}) + c

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e^{x} (\frac{x + 2}{x + 4}) + c

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e^{x} (\frac{x - 2}{x + 4}) + c

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(\frac{2 x e^{2}}{x + 4}) + c

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Solution

The correct option is A $e^{x} (\frac{x}{x + 4}) + c$
$I = \int {(\frac{x + 2}{x + 4})}^{2} e^{x} d x$
$I = \int e^{x} [\frac{x^{2} + 4 x + 4}{(x + 4)^{2}}] d x$
$= \int e^{x} [\frac{x (x + 4)}{(x + 4)^{2}} + \frac{4}{(x + 4)^{2}}] d x$
$= \int e^{x} [\frac{x}{x + 4} + \frac{4}{(x + 4)^{2}}] d x$
$= \int e^{x} [f (x) + f^{'} (x)] d x,$
where $f (x) = \frac{x}{x + 4}$
and $f^{'} (x) = \frac{4}{(x + 4)^{2}}$
$= e^{x} (\frac{x}{x + 4}) + C$

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The correct option is A ex(xx+4)+cI=∫(x+2x+4)2exdxI=∫ex[x2+4x+4(x+4)2]dx=∫ex[x(x+4)(x+4)2+4(x+4)2]dx=∫ex[xx+4+4(x+4)2]dx=∫ex[f(x)+f′(x)]dx,where f(x)=xx+4and f′(x)=4(x+4)2=ex(xx+4)+C

$\int {(\frac{x + 2}{x + 4})}^{2} e^{x} d x$ is equal to