How do you integrate int xe^(-4x) by integration by parts method?

Nov 20, 2016

I found: $= {e}^{- 4 x} / \left(- 4\right) \cdot x - \frac{1}{16} {e}^{4 x} + c =$

Explanation:

If you know about integration by parts you will know that first we integrate one of the functions leaving the other untouched and then subtract the integral of the integrated funtion times the other function derived as:
$\int f \left(x\right) g \left(x\right) \mathrm{dx} = F \left(x\right) g \left(x\right) - \int F \left(x\right) g ' \left(x\right) \mathrm{dx}$
in our case:
$\int x {e}^{- 4 x} \mathrm{dx} = {e}^{- 4 x} / \left(- 4\right) \cdot x - \int \left({e}^{- 4 x} / \left(- 4\right)\right) \cdot 1 \mathrm{dx} =$
$\int x {e}^{- 4 x} \mathrm{dx} = {e}^{- 4 x} / \left(- 4\right) \cdot x + \frac{1}{4} \int {e}^{- 4 x} \mathrm{dx} =$
$= {e}^{- 4 x} / \left(- 4\right) \cdot x + \frac{1}{4} \frac{{e}^{4 x}}{- 4} + c =$
$= {e}^{- 4 x} / \left(- 4\right) \cdot x - \frac{1}{16} {e}^{4 x} + c =$