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

The integral of xye^{-yx} is -e^{-yx}x-(e^{-yx})/y+C

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integral of xye^{-yx}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= y \cdot \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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - y x

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

Simplify

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

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

Add a constant to the solution

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

x^{2} + x y - y^{2} = - 11, (1, 4)

\frac{d}{d x} (\frac{- 2 x ^{2}}{x ^{2} - 1})

2 cot (\frac{x}{8})

x \to - 2 lim (4)

\int \frac{1}{x ^{2} - 225} d x

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