What is the integral of r^3e^{-2r} ?

The integral of r^3e^{-2r} is 1/16 (-8e^{-2r}r^3-3(4e^{-2r}r^2-2(-2e^{-2r}r-e^{-2r})))+C

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integral of r^3e^{-2r}

\int r^{3} e^{- 2 r} d r

\frac{1}{16} (- 8 e^{- 2 r} r^{3} - 3 (4 e^{- 2 r} r^{2} - 2 (- 2 e^{- 2 r} r - e^{- 2 r}))) + C

Solution steps

\int r^{3} e^{- 2 r} d r

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{16} d u

= \int \frac{e ^{u} u ^{3}}{16} d u

Take the constant out:

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

= \frac{1}{16} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= \frac{1}{16} (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= \frac{1}{16} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - 2 r

= \frac{1}{16} ((- 2 r)^{3} e^{- 2 r} - 3 ((- 2 r)^{2} e^{- 2 r} - 2 (e^{- 2 r} (- 2 r) - e^{- 2 r})))

Simplify

\frac{1}{16} ((- 2 r)^{3} e^{- 2 r} - 3 ((- 2 r)^{2} e^{- 2 r} - 2 (e^{- 2 r} (- 2 r) - e^{- 2 r}))) : \frac{1}{16} (- 8 e^{- 2 r} r^{3} - 3 (4 e^{- 2 r} r^{2} - 2 (- 2 e^{- 2 r} r - e^{- 2 r})))

= \frac{1}{16} (- 8 e^{- 2 r} r^{3} - 3 (4 e^{- 2 r} r^{2} - 2 (- 2 e^{- 2 r} r - e^{- 2 r})))

Add a constant to the solution

= \frac{1}{16} (- 8 e^{- 2 r} r^{3} - 3 (4 e^{- 2 r} r^{2} - 2 (- 2 e^{- 2 r} r - e^{- 2 r}))) + C

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

\int_{1}^{2} \frac{9 ln ( x )}{x ^{2}} d x

f (x) = x^{2} - \frac{1}{x}, at x = 2

\frac{d}{d y} (e^{\frac{y}{x}})

y^{^{'}} = \frac{( e ^{x} )}{y}

What is the integral of r^3e^{-2r} ?
The integral of r^3e^{-2r} is 1/16 (-8e^{-2r}r^3-3(4e^{-2r}r^2-2(-2e^{-2r}r-e^{-2r})))+C