What is the integral of 1/2 (x+y)e^{-x-y} ?

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

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integral of 1/2 (x+y)e^{-x-y}

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

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

Solution steps

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

Take the constant out:

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

= \frac{1}{2} \cdot \int (y + x) e^{- y - x} d x

(y + x) e^{- y - x} = e^{- y} (y + x) e^{- x}

= \frac{1}{2} \cdot \int e^{- y} (y + x) e^{- x} d x

Take the constant out:

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

= \frac{1}{2} e^{- y} \cdot \int (y + x) e^{- x} d x

Apply Integration By Parts

: - e^{- x} (y + x) - \int - e^{- x} d x

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

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{1}{2} x^{\frac{4}{5}})

x \to 3 - lim (3 + x)

\frac{d}{d x} (\frac{4}{e ^{x} + e ^{- x}})

\frac{d}{d x} (\frac{x + 2}{5 x ^{2} e ^{x}})

f (x) = 5 ln (x)

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