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

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

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integral of sqrt(x)e^{-sqrt(x)}

\int x e^{- x} d x

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

Solution steps

\int x e^{- x} d x

Apply u-substitution

: \int - 2 e^{u} u^{2} d u

= \int - 2 e^{u} u^{2} d u

Take the constant out:

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

= - 2 \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

: u^{2} e^{u} - \int 2 u e^{u} d u

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

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

= - 2 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - x

= - 2 ((- x)^{2} e^{- x} - 2 (e^{- x} (- x) - e^{- x}))

Simplify

- 2 ((- x)^{2} e^{- x} - 2 (e^{- x} (- x) - e^{- x})) : - 2 (e^{- x} x - 2 (- e^{- x} x - e^{- x}))

= - 2 (e^{- x} x - 2 (- e^{- x} x - e^{- x}))

Add a constant to the solution

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

\frac{d}{d x} (\frac{2 x}{ln ( x )})

\frac{d}{d x} (\frac{x ^{3}}{1 - x ^{3}})

\frac{\partial}{\partial x} (x + e^{z} + y)

\int 4 sin (2 x) - 3 cos (3 x) d x

f (x) = x^{4} + 5 x - 2, (1, 4)

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