What is the integral of 4x^3e^{-5x} ?

The integral of 4x^3e^{-5x} is 4(-1/5 e^{-5x}x^3-3/625 (25e^{-5x}x^2-2(-5e^{-5x}x-e^{-5x})))+C

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integral of 4x^3e^{-5x}

\int 4 x^{3} e^{- 5 x} d x

4 (- \frac{1}{5} e^{- 5 x} x^{3} - \frac{3}{625} (25 e^{- 5 x} x^{2} - 2 (- 5 e^{- 5 x} x - e^{- 5 x}))) + C

Solution steps

\int 4 x^{3} e^{- 5 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int x^{3} e^{- 5 x} d x

Apply Integration By Parts

: - \frac{1}{5} e^{- 5 x} x^{3} - \int - \frac{3}{5} e^{- 5 x} x^{2} d x

= 4 (- \frac{1}{5} e^{- 5 x} x^{3} - \int - \frac{3}{5} e^{- 5 x} x^{2} d x)

\int - \frac{3}{5} e^{- 5 x} x^{2} d x = \frac{3}{625} (25 e^{- 5 x} x^{2} - 2 (- 5 e^{- 5 x} x - e^{- 5 x}))

= 4 (- \frac{1}{5} e^{- 5 x} x^{3} - \frac{3}{625} (25 e^{- 5 x} x^{2} - 2 (- 5 e^{- 5 x} x - e^{- 5 x})))

Add a constant to the solution

= 4 (- \frac{1}{5} e^{- 5 x} x^{3} - \frac{3}{625} (25 e^{- 5 x} x^{2} - 2 (- 5 e^{- 5 x} x - e^{- 5 x}))) + C

\frac{1}{( s + 1 ) ( s ^{2} + 2 s + 10 )}

\frac{d}{d x} (\frac{9 x ^{2} + 4}{x ^{2} + 1})

\int \frac{( x ^{2} + 3 )}{x ^{2} - 3 x + 2} d x

\int x (3 x - 2) d x

\frac{\partial}{\partial y} (2 x^{2})

What is the integral of 4x^3e^{-5x} ?
The integral of 4x^3e^{-5x} is 4(-1/5 e^{-5x}x^3-3/625 (25e^{-5x}x^2-2(-5e^{-5x}x-e^{-5x})))+C