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

The integral of x/(e^{-3x)} is 1/9 (e^{3x}*3x-e^{3x})+C

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

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

\frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + C

Solution steps

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

Simplify

\frac{x}{e ^{- 3 x}} : e^{3 x} x

= \int e^{3 x} x d x

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

= \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 x

= \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x})

Add a constant to the solution

= \frac{1}{9} (e^{3 x} \cdot 3 x - e^{3 x}) + C

\int tan^{2} (w) d w

\frac{\partial}{\partial x} (- a e^{k z} w sin (k x - wt))

y^{^{''}} - 5 y^{^{'}} = 25 t, y (0) = 5, y^{^{'}} (0) = 0

\int_{0}^{2} x 4 - x^{2} d x

\int \frac{arctan ( x )}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

What is the integral of x/(e^{-3x)} ?
The integral of x/(e^{-3x)} is 1/9 (e^{3x}*3x-e^{3x})+C