Question

$\int \frac{(x+3) e^x}{(x+4)^2} d x$ is equal to

• A. $\frac{1}{(x + 4)^2} +C$
• B. $\frac{e^{x}}{(x + 4)^2} +C$
• C. $\frac{e^{x}}{(x + 4)} +C$
• D. $\frac{e^{x}}{(x + 3)} +C$

Answer 1

Answer: (C)

Let $I=\int \frac{x+3}{(x+4)^{2}} e^{x} d x$
$=\int e^{x}\left[\frac{x+4-1}{(x+4)^{2}}\right] d x$
$=\int e^{x}\left[\frac{1}{x+4}+\frac{-1}{(x+4)^{2}}\right] d x$
$=\int e^{x} \frac{1}{x+4} d x+\int e^{x}\left(\frac{-1}{(x+4)^{2}}\right) d x$
$=\frac{1}{x+4} e^{x}-\int e^{x} \cdot\left(\frac{-1}{(x+4)^{2}}\right) d x+$
$\int e^{x}\left(-\frac{1}{(x+4)^{2}}\right) d x$
$=\frac{e^{x}}{x+4}+C$