$\int {(\frac{x + 2}{x + 4})}^{2} e^{x} d x$ 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 $e^{x} (\frac{x}{x + 4}) + C$
$I = \int {(\frac{x + 2}{x + 4})}^{2} e^{x} d x = \int e^{x} [\frac{x^{2} + 4 x + 4}{{(x + 4)}^{2}}] d x = \int e^{x} [\frac{x (x + 4)}{{(x + 4)}^{2}} + \frac{4}{{(x + 4)}^{2}}] d x$
$= \int e^{x} [\frac{x}{x + 4} + \frac{4}{{(x + 4)}^{2}}] d x ↓ f (x) ↓ f^{'} (x) {∵ \int e^{x} {f (x) + f^{'} (x)} d x = e^{x} f (x) + C} I = e^{x} (\frac{x}{x + 4}) + C$

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