What is the integral of x^3e^{7x} ?

The integral of x^3e^{7x} is 1/2401 (343e^{7x}x^3-3(49e^{7x}x^2-2(7e^{7x}x-e^{7x})))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of x^3e^{7x}

\int x^{3} e^{7 x} d x

\frac{1}{2401} (343 e^{7 x} x^{3} - 3 (49 e^{7 x} x^{2} - 2 (7 e^{7 x} x - e^{7 x}))) + C

Solution steps

\int x^{3} e^{7 x} d x

Apply u-substitution

= \int \frac{e ^{u} u ^{3}}{2401} d u

Take the constant out:

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

= \frac{1}{2401} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

= \frac{1}{2401} (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}))

= \frac{1}{2401} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = 7 x

= \frac{1}{2401} ((7 x)^{3} e^{7 x} - 3 ((7 x)^{2} e^{7 x} - 2 (e^{7 x} \cdot 7 x - e^{7 x})))

Simplify

\frac{1}{2401} ((7 x)^{3} e^{7 x} - 3 ((7 x)^{2} e^{7 x} - 2 (e^{7 x} \cdot 7 x - e^{7 x}))) : \frac{1}{2401} (343 e^{7 x} x^{3} - 3 (49 e^{7 x} x^{2} - 2 (7 e^{7 x} x - e^{7 x})))

= \frac{1}{2401} (343 e^{7 x} x^{3} - 3 (49 e^{7 x} x^{2} - 2 (7 e^{7 x} x - e^{7 x})))

Add a constant to the solution

= \frac{1}{2401} (343 e^{7 x} x^{3} - 3 (49 e^{7 x} x^{2} - 2 (7 e^{7 x} x - e^{7 x}))) + C

y^{^{'}} - 3 y = 6, y (0) = 1

\int e^{a x} x d x

\frac{d}{d x} (5 g (x) (2 x^{3}) + \frac{2}{x ^{2}})

\int_{0}^{3} e^{- 2 x} d x

\frac{\partial}{\partial x} (2 x y e^{- x^{2} y})

What is the integral of x^3e^{7x} ?
The integral of x^3e^{7x} is 1/2401 (343e^{7x}x^3-3(49e^{7x}x^2-2(7e^{7x}x-e^{7x})))+C