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

The integral of x^3e^{-ax} is 1/(a^4)(-a^3e^{-ax}x^3-3(a^2e^{-ax}x^2-2(-ae^{-ax}x-e^{-ax})))+C

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integral of x^3e^{-ax}

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

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

Solution steps

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

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{a ^{4}} d u

= \int \frac{e ^{u} u ^{3}}{a ^{4}} d u

Take the constant out:

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

= \frac{1}{a ^{4}} \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}{a ^{4}} (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}{a ^{4}} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - a x

= \frac{1}{a ^{4}} ((- a x)^{3} e^{- a x} - 3 ((- a x)^{2} e^{- a x} - 2 (e^{- a x} (- a x) - e^{- a x})))

Simplify

\frac{1}{a ^{4}} ((- a x)^{3} e^{- a x} - 3 ((- a x)^{2} e^{- a x} - 2 (e^{- a x} (- a x) - e^{- a x}))) : \frac{1}{a ^{4}} (- a^{3} e^{- a x} x^{3} - 3 (a^{2} e^{- a x} x^{2} - 2 (- a e^{- a x} x - e^{- a x})))

= \frac{1}{a ^{4}} (- a^{3} e^{- a x} x^{3} - 3 (a^{2} e^{- a x} x^{2} - 2 (- a e^{- a x} x - e^{- a x})))

Add a constant to the solution

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

\int (3 x - 1)^{2003} d x

\int \frac{1}{x - 3 x + 2} d x

\int e^{- 0.08 t} d t

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