Question
$\int e ^x\left[\frac{(x+3)}{(x+4)^2}\right] d x$
✓
Answer
$\int e ^x\left[\frac{(x+3)}{(x+4)^2}\right] d x=\int e ^x\left[\frac{x+4-1}{(x+4)^2}\right] d x$
$=\int e ^x\left[\frac{1}{x+4}-\frac{1}{(x+4)^2}\right] d x$
$= e ^x\left(\frac{1}{x+4}\right)+ c \quad \ldots \ldots . .\left[\because e ^x\left[ f (x)+ f ^{\prime}(x)\right] d x= e ^x \cdot f (x)+ c \right]$
$=\int e ^x\left[\frac{1}{x+4}-\frac{1}{(x+4)^2}\right] d x$
$= e ^x\left(\frac{1}{x+4}\right)+ c \quad \ldots \ldots . .\left[\because e ^x\left[ f (x)+ f ^{\prime}(x)\right] d x= e ^x \cdot f (x)+ c \right]$
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