What is the integral of x^3e^{-4x} ?

The integral of x^3e^{-4x} is -1/4 e^{-4x}x^3-3/128 e^{-4x}(8x^2+4x+1)+C

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integral of x^3e^{-4x}

\int x^{3} e^{- 4 x} d x

- \frac{1}{4} e^{- 4 x} x^{3} - \frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1) + C

Solution steps

\int x^{3} e^{- 4 x} d x

Apply Integration By Parts

: - \frac{1}{4} e^{- 4 x} x^{3} - \int - \frac{3}{4} e^{- 4 x} x^{2} d x

= - \frac{1}{4} e^{- 4 x} x^{3} - \int - \frac{3}{4} e^{- 4 x} x^{2} d x

\int - \frac{3}{4} e^{- 4 x} x^{2} d x = \frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1)

= - \frac{1}{4} e^{- 4 x} x^{3} - \frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1)

Add a constant to the solution

= - \frac{1}{4} e^{- 4 x} x^{3} - \frac{3}{128} e^{- 4 x} (8 x^{2} + 4 x + 1) + C

(1 + x^{2}) y^{^{'}} - 2 x y = x^{2} + x^{4}

y = x^{4} + x^{3}, y = x^{4} + x^{2} + 6 x

\frac{d}{d x} (e^{x} sin (4 x))

\frac{d}{d x} (A e^{4 x})

\frac{\partial}{\partial v} (2 u + v)

What is the integral of x^3e^{-4x} ?
The integral of x^3e^{-4x} is -1/4 e^{-4x}x^3-3/128 e^{-4x}(8x^2+4x+1)+C