What is the integral of xe^{-yx} ?

The integral of xe^{-yx} is 1/(y^2)(-ye^{-yx}x-e^{-yx})+C

Popular >

integral of xe^{-yx}

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

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

Solution steps

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

Apply u-substitution

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

= \int \frac{e ^{u} u}{y ^{2}} d u

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - y x

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

Simplify

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

Add a constant to the solution

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

\frac{d}{d x} (sin^{2} (x^{2} + 1))

\int_{1}^{4} - 5 t ln (t) d t

\frac{\partial}{\partial x} (x^{2} cos (x y^{2}))

t \to + 0 lim (\frac{e ^{t} - 1}{t ^{9}})

f (x) = \frac{3}{x}, (6, \frac{1}{2})

What is the integral of xe^{-yx} ?
The integral of xe^{-yx} is 1/(y^2)(-ye^{-yx}x-e^{-yx})+C