What is the integral of-6sqrt(x)e^{sqrt(x)} ?

The integral of-6sqrt(x)e^{sqrt(x)} is -12(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C

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integral of-6sqrt(x)e^{sqrt(x)}

\int - 6 x e^{x} d x

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

Solution steps

\int - 6 x e^{x} d x

Take the constant out:

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

= - 6 \cdot \int x e^{x} d x

Apply u-substitution

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

= - 6 \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

= - 6 \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

= - 6 \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})

= - 6 \cdot 2 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = x

= - 6 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x}))

Simplify

- 6 \cdot 2 ((x)^{2} e^{x} - 2 (e^{x} x - e^{x})) : - 12 (e^{x} x - 2 (e^{x} x - e^{x}))

= - 12 (e^{x} x - 2 (e^{x} x - e^{x}))

Add a constant to the solution

= - 12 (e^{x} x - 2 (e^{x} x - e^{x})) + C

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

\frac{d y}{d x} = 3 (2 - y)

\frac{d y}{d x} = \frac{x y + y ^{2}}{x ^{2}}

\frac{d y}{d x} = 16 4 4 x^{4} + 4

y^{^{''}} + 2 y^{^{'}} + 26 y = 0

What is the integral of-6sqrt(x)e^{sqrt(x)} ?
The integral of-6sqrt(x)e^{sqrt(x)} is -12(e^{sqrt(x)}x-2(e^{sqrt(x)}sqrt(x)-e^{sqrt(x)}))+C