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

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

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

integral

x^{3} (e^{- x})^{4}

- \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^{- x})^{4} d x

(e^{- x})^{4} = e^{- 4 x}

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

- t + 3

(arccos (x))^{2}

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

\frac{1}{( v + 1 ) ^{2}}

3 t^{2} + 2 t + 1

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