What is the integral of 6te^{3t} ?

The integral of 6te^{3t} is 2e^{3t}t-2/3 e^{3t}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 6te^{3t}

\int 6 t e^{3 t} d t

2 e^{3 t} t - \frac{2}{3} e^{3 t} + C

Solution steps

\int 6 t e^{3 t} d t

Take the constant out:

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

= 6 \cdot \int t e^{3 t} d t

Apply u-substitution

: \int \frac{e ^{u} u}{9} d u

= 6 \cdot \int \frac{e ^{u} u}{9} d u

Take the constant out:

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

= 6 \cdot \frac{1}{9} \cdot \int e^{u} u d u

Apply Integration By Parts

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

= 6 \cdot \frac{1}{9} (e^{u} u - \int e^{u} d u)

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

= 6 \cdot \frac{1}{9} (e^{u} u - e^{u})

Substitute back

u = 3 t

= 6 \cdot \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t})

Simplify

6 \cdot \frac{1}{9} (e^{3 t} \cdot 3 t - e^{3 t}) : 2 e^{3 t} t - \frac{2}{3} e^{3 t}

= 2 e^{3 t} t - \frac{2}{3} e^{3 t}

Add a constant to the solution

= 2 e^{3 t} t - \frac{2}{3} e^{3 t} + C

x \to 0 lim (\frac{x ^{3}}{x ^{3}})

\frac{d}{d x} (a^{2} sin^{2} (x))

f (x) = \frac{x}{x + 3}, at x = - 2

t \to + 0 lim (\frac{sin ( 2 t )}{t})

\frac{( s + 2 )}{( s ^{2} + 4 s + 13 )}

What is the integral of 6te^{3t} ?
The integral of 6te^{3t} is 2e^{3t}t-2/3 e^{3t}+C