Question asks to show that if $f(x)= \begin{cases} \frac14xe^\frac{-x}{2} & x>0\\[8pt] 0 & \text{elsewhere}, \end{cases}$ then $\int_0^\infty f(x)\,dx=1.$
I get
$\int f(x) = \frac14 \left [-4e^\frac{-x}{2}(-\frac{x}{2} - 1) + C \right] $ And I don't get how this ends up being equal to 1.