What is the integral of 8xe^{0.5x^2} ?

The integral of 8xe^{0.5x^2} is 8e^{0.5x^2}+C

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integral of 8xe^{0.5x^2}

\int 8 x e^{0.5 x^{2}} d x

8 e^{0.5 x^{2}} + C

Solution steps

\int 8 x e^{0.5 x^{2}} d x

Take the constant out:

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

= 8 \cdot \int x e^{0.5 x^{2}} d x

Apply u-substitution

: \int e^{u} d u

= 8 \cdot \int e^{u} d u

Use the common integral:

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

= 8 e^{u}

Substitute back

u = 0.5 x^{2}

= 8 e^{0.5 x^{2}}

Add a constant to the solution

= 8 e^{0.5 x^{2}} + C

3 y^{^{''}} + 108 y = 0

\frac{\partial}{\partial x} ((x - y) (x y - 1))

\frac{d}{d x} (\frac{x \cdot sin ( x )}{1 + cos ( x )})

\int (x - 3) (x + 3) d x

\frac{d y}{d x} - \frac{1 y}{x} = 1 - \frac{1}{x}

What is the integral of 8xe^{0.5x^2} ?
The integral of 8xe^{0.5x^2} is 8e^{0.5x^2}+C