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}))

(x)^{2} = 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

2 x y^{^{'}} + y = 6 x, y (4) = 26

f (x) = \frac{1}{x}

\int (x + 2) ln (x + 2) 2 d x

\int_{0}^{4} 5 x - x^{2} d x

\int e^{u} u d u

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