Find the value of the integral $\int \frac{(x+3)}{(x+4)^{2}} e^{x} \: dx$

 For finding the value of the integral $\int \frac{(x+3)}{(x+4)^{2}} e^{x} \: dx$  , consider the property of indefinite integral:

\[\int e^{x} [f(x) + {f}'(x)] dx= e^{x}f(x)+C\]

Consider the function:

\[\begin{align*} \left [\frac{x+3}{(x+4)^{2}} \right ] e^{x} & =\left [ \frac{x+3+1-1}{(x+4)^{2}} \right ]e^{x} \\ &= \left [ \frac{x+4}{(x+4)^{2}} - \frac{1}{(x+4)^{2}} \right ] e^{x}\\ &=\left [ \frac{1}{x+4} - \frac{1}{(x+4)^{2}}\right ] e^{x}\\ &= \left [ f(x)+{f}'(x) \right ] e^{x} \end{align*}\]

where 

\[f(x)=\frac{1}{x+4} \; and\; {f}'(x)=\frac{\mathrm{d}f }{\mathrm{d} x}=\frac{-1}{(x+4)^{2}}\]

So by the property of definite integral given above

\[\begin{align*} \int e^{x}\left [ f(x) +{f}'(x) \right ] dx &= \int e^{x}\left [ \frac{1}{(x+4)}-\frac{1}{(x+4)^{2}} \right ] dx\\ &= e^{x}\left [\frac{1}{x+4} \right ]+C\; \; \left (=e^{x}f(x)+C \right ) \end{align*}\]

\[\therefore \int e^{x}\left [\frac{x+3}{(x+4)^{2}} \right ] dx = e^{x}\left [\frac{1}{x+4} \right ]+C\]

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