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

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

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integral of e^{-ax}x

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

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

Solution steps

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

Apply u-substitution

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

= \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - a x

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

Simplify

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

Add a constant to the solution

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

\int_{0}^{1} \frac{- 5 x}{e ^{2 x}} d x

\frac{d}{d x} (16 (x + 2)^{3})

\frac{2}{5}

f (x) = \frac{3 x}{7 x ^{2} + 1}

\frac{\partial}{\partial x} (3 y - \frac{x}{y ^{2}})

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