What is the integral of 20xe^{8x} ?

The integral of 20xe^{8x} is 5/16 (8e^{8x}x-e^{8x})+C

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integral of 20xe^{8x}

\int 20 x e^{8 x} d x

\frac{5}{16} (8 e^{8 x} x - e^{8 x}) + C

Solution steps

\int 20 x e^{8 x} d x

Take the constant out:

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

= 20 \cdot \int x e^{8 x} d x

Apply u-substitution

: \int \frac{e ^{u} u}{64} d u

= 20 \cdot \int \frac{e ^{u} u}{64} d u

Take the constant out:

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

= 20 \cdot \frac{1}{64} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 20 \cdot \frac{1}{64} (e^{u} u - \int e^{u} d u)

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

= 20 \cdot \frac{1}{64} (e^{u} u - e^{u})

Substitute back

u = 8 x

= 20 \cdot \frac{1}{64} (e^{8 x} \cdot 8 x - e^{8 x})

Simplify

20 \cdot \frac{1}{64} (e^{8 x} \cdot 8 x - e^{8 x}) : \frac{5}{16} (8 e^{8 x} x - e^{8 x})

= \frac{5}{16} (8 e^{8 x} x - e^{8 x})

Add a constant to the solution

= \frac{5}{16} (8 e^{8 x} x - e^{8 x}) + C

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What is the integral of 20xe^{8x} ?
The integral of 20xe^{8x} is 5/16 (8e^{8x}x-e^{8x})+C