What is the integral of (e^{3x}-4)^5*3e^{3x} ?

The integral of (e^{3x}-4)^5*3e^{3x} is 1/6 (e^{3x}-4)^6+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (e^{3x}-4)^5*3e^{3x}

\int (e^{3 x} - 4)^{5} \cdot 3 e^{3 x} d x

\frac{1}{6} (e^{3 x} - 4)^{6} + C

Solution steps

\int (e^{3 x} - 4)^{5} \cdot 3 e^{3 x} d x

Take the constant out:

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

= 3 \cdot \int (e^{3 x} - 4)^{5} e^{3 x} d x

Apply u-substitution

: \int \frac{u ^{5}}{3} d u

= 3 \cdot \int \frac{u ^{5}}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int u^{5} d u

Apply the Power Rule

: \frac{u ^{6}}{6}

= 3 \cdot \frac{1}{3} \cdot \frac{u ^{6}}{6}

Substitute back

u = e^{3 x} - 4

= 3 \cdot \frac{1}{3} \cdot \frac{( e ^{3 x} - 4 ) ^{6}}{6}

Simplify

3 \cdot \frac{1}{3} \cdot \frac{( e ^{3 x} - 4 ) ^{6}}{6} : \frac{1}{6} (e^{3 x} - 4)^{6}

= \frac{1}{6} (e^{3 x} - 4)^{6}

Add a constant to the solution

= \frac{1}{6} (e^{3 x} - 4)^{6} + C

\int arccot (9 y) d y

\frac{d ^{2}}{d x ^{2}} ((x^{3} + 8) e^{x})

\frac{d}{d x} (r (x) x)

x \cdot e^{x}

x \to 2 lim (8)

What is the integral of (e^{3x}-4)^5*3e^{3x} ?
The integral of (e^{3x}-4)^5*3e^{3x} is 1/6 (e^{3x}-4)^6+C