What is the integral of 4e^{x/2}x ?

The integral of 4e^{x/2}x is 16(1/2 e^{x/2}x-e^{x/2})+C

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integral of 4e^{x/2}x

\int 4 e^{\frac{x}{2}} x d x

16 (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}}) + C

Solution steps

\int 4 e^{\frac{x}{2}} x d x

Take the constant out:

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

= 4 \cdot \int e^{\frac{x}{2}} x d x

Apply u-substitution

: \int 4 e^{u} u d u

= 4 \cdot \int 4 e^{u} u d u

Take the constant out:

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

= 4 \cdot 4 \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot 4 (e^{u} u - \int e^{u} d u)

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

= 4 \cdot 4 (e^{u} u - e^{u})

Substitute back

u = \frac{x}{2}

= 4 \cdot 4 (e^{\frac{x}{2}} \frac{x}{2} - e^{\frac{x}{2}})

Simplify

4 \cdot 4 (e^{\frac{x}{2}} \frac{x}{2} - e^{\frac{x}{2}}) : 16 (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}})

= 16 (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}})

Add a constant to the solution

= 16 (\frac{1}{2} e^{\frac{x}{2}} x - e^{\frac{x}{2}}) + C

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What is the integral of 4e^{x/2}x ?
The integral of 4e^{x/2}x is 16(1/2 e^{x/2}x-e^{x/2})+C