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$\int \frac{x+3}{(x+4)^2} e^x d x$ is equal to
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Verified Answer
The correct answer is:
$e^x\left(\frac{1}{x+4}\right)+C$
Suppose,
$I=\int \frac{x+3}{(x+4)^2} e^x d x=\int \frac{(x+4) 1}{(x+4)^2} e^x d x$
$=\int \frac{e^x}{(x+4)}-\int \frac{e^x}{(x+4)^2} d x$
$=\int e^x\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^2}\right) d x$
$=e^x \frac{1}{(x+4)}+C$
$\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C\right]$
$I=\int \frac{x+3}{(x+4)^2} e^x d x=\int \frac{(x+4) 1}{(x+4)^2} e^x d x$
$=\int \frac{e^x}{(x+4)}-\int \frac{e^x}{(x+4)^2} d x$
$=\int e^x\left(\frac{1}{(x+4)}-\frac{1}{(x+4)^2}\right) d x$
$=e^x \frac{1}{(x+4)}+C$
$\left[\int e^x\left\{f(x)+f^{\prime}(x)\right\} d x=e^x f(x)+C\right]$
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