What is the integral of 9x^3e^{3x} ?

The integral of 9x^3e^{3x} is 1/3 e^{3x}(9x^3-9x^2+6x-2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9x^3e^{3x}

\int 9 x^{3} e^{3 x} d x

\frac{1}{3} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

Solution steps

\int 9 x^{3} e^{3 x} d x

Take the constant out:

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

= 9 \cdot \int x^{3} e^{3 x} d x

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{81} d u

= 9 \cdot \int \frac{e ^{u} u ^{3}}{81} d u

Take the constant out:

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

= 9 \cdot \frac{1}{81} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= 9 \cdot \frac{1}{81} (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

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

= 9 \cdot \frac{1}{81} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = 3 x

= 9 \cdot \frac{1}{81} ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x})))

Simplify

9 \cdot \frac{1}{81} ((3 x)^{3} e^{3 x} - 3 ((3 x)^{2} e^{3 x} - 2 (e^{3 x} \cdot 3 x - e^{3 x}))) : \frac{1}{3} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

= \frac{1}{3} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2)

Add a constant to the solution

= \frac{1}{3} e^{3 x} (9 x^{3} - 9 x^{2} + 6 x - 2) + C

n = 1 \sum \infty \frac{1}{3 ^{n} + n}

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

f (x) = (3 x^{6} + 2 x^{3})^{4}

\frac{d y}{d x} = y (3 y - 6)

\frac{d}{d x} (\frac{1}{3 x ^{4} + 1})

What is the integral of 9x^3e^{3x} ?
The integral of 9x^3e^{3x} is 1/3 e^{3x}(9x^3-9x^2+6x-2)+C