What is the integral of 3e^{(3t)/2}t ?

The integral of 3e^{(3t)/2}t is 2e^{(3t)/2}t-4/3 e^{(3t)/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3e^{(3t)/2}t

\int 3 e^{\frac{3 t}{2}} t d t

2 e^{\frac{3 t}{2}} t - \frac{4}{3} e^{\frac{3 t}{2}} + C

Solution steps

\int 3 e^{\frac{3 t}{2}} t d t

Take the constant out:

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

= 3 \cdot \int e^{\frac{3 t}{2}} t d t

Apply Integration By Parts

: \frac{2}{3} e^{\frac{3 t}{2}} t - \int \frac{2}{3} e^{\frac{3 t}{2}} d t

= 3 (\frac{2}{3} e^{\frac{3 t}{2}} t - \int \frac{2}{3} e^{\frac{3 t}{2}} d t)

\int \frac{2}{3} e^{\frac{3 t}{2}} d t = \frac{4}{9} e^{\frac{3 t}{2}}

= 3 (\frac{2}{3} e^{\frac{3 t}{2}} t - \frac{4}{9} e^{\frac{3 t}{2}})

Simplify

3 (\frac{2}{3} e^{\frac{3 t}{2}} t - \frac{4}{9} e^{\frac{3 t}{2}}) : 2 e^{\frac{3 t}{2}} t - \frac{4}{3} e^{\frac{3 t}{2}}

= 2 e^{\frac{3 t}{2}} t - \frac{4}{3} e^{\frac{3 t}{2}}

Add a constant to the solution

= 2 e^{\frac{3 t}{2}} t - \frac{4}{3} e^{\frac{3 t}{2}} + C

f (x) = 4 x + 6 x, at x = 4

f (x) = - 3 - 4 x^{2}, (5, - 103)

f (x) = 2 x^{3} - x^{2} + 1, at x = 2

sin (x) - cos (x)

x \frac{d y}{d x} + y = x^{4} ln (x)

What is the integral of 3e^{(3t)/2}t ?
The integral of 3e^{(3t)/2}t is 2e^{(3t)/2}t-4/3 e^{(3t)/2}+C