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

The integral of 8xe^{8x} is e^{8x}x-1/8 e^{8x}+C

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integral of 8xe^{8x}

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

e^{8 x} x - \frac{1}{8} e^{8 x} + C

Solution steps

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

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{64} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 8 \cdot \frac{1}{64} (e^{u} u - \int e^{u} d u)

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

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

Substitute back

u = 8 x

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

Simplify

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

= e^{8 x} x - \frac{1}{8} e^{8 x}

Add a constant to the solution

= e^{8 x} x - \frac{1}{8} e^{8 x} + C

ln^{3} (x)

\frac{d y}{d x} = \frac{( 1 - 3 y ) ^{2}}{x}, y (1) = 1

\frac{d}{d x} ((x^{2} + 1)^{2})

n = 0 \sum \infty \frac{n ^{2}}{n !}

\frac{d}{d x} (sin (3 x) ln (sech (2 x)))

What is the integral of 8xe^{8x} ?
The integral of 8xe^{8x} is e^{8x}x-1/8 e^{8x}+C