What is the integral of e^{3t}(e^{3t}-1) ?

The integral of e^{3t}(e^{3t}-1) is 1/6 (e^{3t}-1)^2+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of e^{3t}(e^{3t}-1)

\int e^{3 t} (e^{3 t} - 1) d t

\frac{1}{6} (e^{3 t} - 1)^{2} + C

Solution steps

\int e^{3 t} (e^{3 t} - 1) d t

Apply u-substitution

: \int \frac{u}{3} d u

= \int \frac{u}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= \frac{1}{3} \cdot \frac{u ^{2}}{2}

Substitute back

u = e^{3 t} - 1

= \frac{1}{3} \cdot \frac{( e ^{3 t} - 1 ) ^{2}}{2}

Simplify

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

= \frac{1}{6} (e^{3 t} - 1)^{2}

Add a constant to the solution

= \frac{1}{6} (e^{3 t} - 1)^{2} + C

\frac{( s + 1 )}{( s + 1 ) ^{2} + 1}

(- 1, 6), (7, - 2)

\int \frac{1}{10 x + 1} d x

\frac{d y}{d x} = y - x

F (C) = \frac{9}{5} C + 32

What is the integral of e^{3t}(e^{3t}-1) ?
The integral of e^{3t}(e^{3t}-1) is 1/6 (e^{3t}-1)^2+C