What is the integral of x/2 e^{-x/2} ?

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

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integral of x/2 e^{-x/2}

\int \frac{x}{2} e^{- \frac{x}{2}} d x

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

Solution steps

\int \frac{x}{2} e^{- \frac{x}{2}} 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 x e^{- \frac{x}{2}} d x

Apply Integration By Parts

: - 2 e^{- \frac{x}{2}} x - \int - 2 e^{- \frac{x}{2}} d x

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

\int - 2 e^{- \frac{x}{2}} d x = 4 e^{- \frac{x}{2}}

= \frac{1}{2} (- 2 e^{- \frac{x}{2}} x - 4 e^{- \frac{x}{2}})

Simplify

\frac{1}{2} (- 2 e^{- \frac{x}{2}} x - 4 e^{- \frac{x}{2}}) : - e^{- \frac{x}{2}} (x + 2)

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

Add a constant to the solution

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

(1 + x^{2}) \cdot \frac{d y}{d x} - \frac{4 x ^{3} y}{1 - x ^{2}} = 1

\int_{0}^{1} \frac{15 5 x + 4}{2} d x

\frac{d y}{d x} - 7 y = e^{x}

y^{^{'}} = \frac{11 x ^{2} - 1}{11 + 4 y}

h (t) = (t + 1)^{\frac{2}{3}} (2 t^{2} - 5)^{3}

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