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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 4xe^{4x}

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

e^{4 x} x - \frac{1}{4} e^{4 x} + C

Solution steps

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

Apply u-substitution

= 4 \cdot \int \frac{e ^{u} u}{16} d u

Take the constant out:

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

= 4 \cdot \frac{1}{16} \cdot \int e^{u} u d u

Apply Integration By Parts

= 4 \cdot \frac{1}{16} (e^{u} u - \int e^{u} d u)

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

= 4 \cdot \frac{1}{16} (e^{u} u - e^{u})

Substitute back

u = 4 x

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

Simplify

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

= e^{4 x} x - \frac{1}{4} e^{4 x}

Add a constant to the solution

= e^{4 x} x - \frac{1}{4} e^{4 x} + C

\frac{d}{d x} (\frac{x ^{2}}{8} - \frac{1}{x})

\int \frac{1}{3 x ^{2} + 4} d x

t an g e n t y = 3 x + 6 cos (x), (\frac{π}{3}, π + 3)

d er i v a t i v e y = (x^{2} + 8)^{2} - 4 x

\frac{d}{d x} (\frac{x ^{5}}{10})

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