What is the integral of xae^{-ax} ?

The integral of xae^{-ax} is -e^{-ax}x-(e^{-ax})/a+C

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integral of xae^{-ax}

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

- e^{- a x} x - \frac{e ^{- a x}}{a} + C

Solution steps

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

Take the constant out:

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

= a \cdot \int x e^{- a x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{a ^{2}} d u

= a \cdot \int \frac{e ^{u} u}{a ^{2}} d u

Take the constant out:

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

= a \frac{1}{a ^{2}} \cdot \int e^{u} u d u

Apply Integration By Parts

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

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

\int e^{u} d u = e^{u}

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

Substitute back

u = - a x

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

Simplify

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

= - e^{- a x} x - \frac{e ^{- a x}}{a}

Add a constant to the solution

= - e^{- a x} x - \frac{e ^{- a x}}{a} + C

y^{^{'}} + 3 t y = t, y (0) = 1

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

x \to 0 lim (x (cot (2 x)))

x, x^{2}

\frac{d}{d x} (1 + x^{3})

What is the integral of xae^{-ax} ?
The integral of xae^{-ax} is -e^{-ax}x-(e^{-ax})/a+C