What is the integral of-x^3e^x ?

The integral of-x^3e^x is -e^xx^3+3e^xx^2-6e^xx+6e^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^3e^x

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

- e^{x} x^{3} + 3 e^{x} x^{2} - 6 e^{x} x + 6 e^{x} + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Simplify

- (x^{3} e^{x} - 3 (x^{2} e^{x} - 2 (e^{x} x - e^{x}))) : - e^{x} x^{3} + 3 e^{x} x^{2} - 6 e^{x} x + 6 e^{x}

= - e^{x} x^{3} + 3 e^{x} x^{2} - 6 e^{x} x + 6 e^{x}

Add a constant to the solution

= - e^{x} x^{3} + 3 e^{x} x^{2} - 6 e^{x} x + 6 e^{x} + C

\int \frac{sec ^{4} ( x )}{sec ^{2} ( x ) - 1} d x

\int \frac{4 z}{5 z ^{2} + 2} d z

\int sin (A x) d x

\int 3 (3 x + 9)^{4} d x

\int (3 - 4 x) (4 x^{2} - 6 x + 7)^{10} d x

What is the integral of-x^3e^x ?
The integral of-x^3e^x is -e^xx^3+3e^xx^2-6e^xx+6e^x+C