Hi:
Would somebody check this for me please.
Also, is this simplified correctly?
thanks
$\displaystyle \int u dv = uv- \int v du $
$\displaystyle \int 9xe^{-4x}= 9\int xe^{-4x} $
u=x, du=dx and dv=e^{-4x} , v=(-1/4)e^{-4x}
$\displaystyle 9\int xe^{-4x}= 9[-\frac{1}{4}e^{-4x}x-\int -\frac{1}{4}e^{-4x}dx] $
$\displaystyle = 9[-\frac{1}{4}e^{-4x}x +\frac{1}{4}\int e^{-4x}dx] $
$\displaystyle = 9[-\frac{1}{4}e^{-4x}x +\frac{1}{4}(e^{-4x}(-\frac{1}{4})]+C $
$\displaystyle = 9[-\frac{1}{4}e^{-4x}x -\frac{1}{8}(e^{-4x})]+C $