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\( \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x \) is eqal to
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Verified Answer
The correct answer is:
\( \frac{e^{x}}{(x+4)}+C \)
Given that \( \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x \)
\( =\int e^{x} \frac{(x+4-1)}{(x+4)^{2} d x} \)
\( =\int e^{x}\left[\frac{1}{x+4}-\frac{1}{(x+4)^{2}}\right] d x \)
Since,
\( \int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c . \)
So, \( \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x=\frac{e^{x}}{x+4}+c \)
\( =\int e^{x} \frac{(x+4-1)}{(x+4)^{2} d x} \)
\( =\int e^{x}\left[\frac{1}{x+4}-\frac{1}{(x+4)^{2}}\right] d x \)
Since,
\( \int e^{x}\left(f(x)+f^{\prime}(x)\right) d x=e^{x} f(x)+c . \)
So, \( \int \frac{(x+3) e^{x}}{(x+4)^{2}} d x=\frac{e^{x}}{x+4}+c \)
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