What is the integral of 40sqrt(x)e^{sqrt(x)} ?

The integral of 40sqrt(x)e^{sqrt(x)} is 80(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C

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integral of 40sqrt(x)e^{sqrt(x)}

\int 40 x e^{x} d x

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

Solution steps

\int 40 x e^{x} d x

Take the constant out:

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

= 40 \cdot \int x e^{x} d x

Apply u-substitution

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

= 40 \cdot \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

= 40 \cdot 2 \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= 40 \cdot 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})

= 40 \cdot 2 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = x

= 40 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x}))

Simplify

40 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x})) : 80 (e^{x} x - 2 (e^{x} x - e^{x}))

= 80 (e^{x} x - 2 (e^{x} x - e^{x}))

Add a constant to the solution

= 80 (e^{x} x - 2 (e^{x} x - e^{x})) + C

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What is the integral of 40sqrt(x)e^{sqrt(x)} ?
The integral of 40sqrt(x)e^{sqrt(x)} is 80(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C