What is the integral of 4xe^{8x} ?

The integral of 4xe^{8x} is 1/16 (8e^{8x}x-e^{8x})+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4xe^{8x}

\int 4 x e^{8 x} d x

\frac{1}{16} (8 e^{8 x} x - e^{8 x}) + C

Solution steps

\int 4 x e^{8 x} d x

Take the constant out:

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

= 4 \cdot \int x e^{8 x} d x

Apply u-substitution

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

= 4 \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

= 4 \cdot \frac{1}{64} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 4 \cdot \frac{1}{64} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{1}{64} (e^{u} u - e^{u})

Substitute back

u = 8 x

= 4 \cdot \frac{1}{64} (e^{8 x} \cdot 8 x - e^{8 x})

Simplify

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

= \frac{1}{16} (8 e^{8 x} x - e^{8 x})

Add a constant to the solution

= \frac{1}{16} (8 e^{8 x} x - e^{8 x}) + C

y^{^{''}} + 6 y^{^{'}} + 5 y = 0

x \to 0 lim (x^{x})

\frac{1}{2} x^{2} 16 - x^{2}

y^{^{''}} - 2 y^{^{'}} + 3 y = 6 sin (3 x) - 6 cos (3 x)

(10 + x^{2}) y^{^{'}} - 2 x y = 0

What is the integral of 4xe^{8x} ?
The integral of 4xe^{8x} is 1/16 (8e^{8x}x-e^{8x})+C