What is the integral of (x^3+3x+8)e^x ?

The integral of (x^3+3x+8)e^x is e^xx^3-3e^xx^2+9e^xx-e^x+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3+3x+8)e^x

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

e^{x} x^{3} - 3 e^{x} x^{2} + 9 e^{x} x - e^{x} + C

Solution steps

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

Expand

(x^{3} + 3 x + 8) e^{x} : e^{x} x^{3} + 3 e^{x} x + 8 e^{x}

= \int e^{x} x^{3} + 3 e^{x} x + 8 e^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{x} x^{3} d x + \int 3 e^{x} x d x + \int 8 e^{x} d x

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

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

\int 8 e^{x} d x = 8 e^{x}

= x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x})) + 3 (e^{x} x - e^{x}) + 8 e^{x}

Simplify

x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x})) + 3 (e^{x} x - e^{x}) + 8 e^{x} : e^{x} x^{3} - 3 e^{x} x^{2} + 9 e^{x} x - e^{x}

= e^{x} x^{3} - 3 e^{x} x^{2} + 9 e^{x} x - e^{x}

Add a constant to the solution

= e^{x} x^{3} - 3 e^{x} x^{2} + 9 e^{x} x - e^{x} + C

x \to - \infty lim (x^{6} e^{x})

\frac{d}{d x} (ln (e^{- x}))

\frac{1}{s + 1}

\int \frac{y - 2}{y ( y - 4 ) ( y + 3 )} d y

\frac{( x ^{2} - 4 )}{x}

What is the integral of (x^3+3x+8)e^x ?
The integral of (x^3+3x+8)e^x is e^xx^3-3e^xx^2+9e^xx-e^x+C