What is the integral of 2x^2ye^{xyz} ?

The integral of 2x^2ye^{xyz} is (2(y^2z^2e^{yzx}x^2-2(yze^{yzx}x-e^{yzx})))/(y^2z^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 2x^2ye^{xyz}

\int 2 x^{2} y e^{x yz} d x

\frac{2 ( y ^{2} z ^{2} e ^{yz x} x ^{2} - 2 ( yz e ^{yz x} x - e ^{yz x} ) )}{y ^{2} z ^{3}} + C

Solution steps

\int 2 x^{2} y e^{x yz} d x

Take the constant out:

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

= 2 y \cdot \int e^{yz x} x^{2} d x

Apply u-substitution

: \int \frac{e ^{u} u ^{2}}{y ^{3} z ^{3}} d u

= 2 y \cdot \int \frac{e ^{u} u ^{2}}{y ^{3} z ^{3}} d u

Take the constant out:

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

= 2 y \frac{1}{y ^{3} z ^{3}} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= 2 y \frac{1}{y ^{3} z ^{3}} (u^{2} e^{u} - \int 2 u e^{u} d u)

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

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

Substitute back

u = yz x

= 2 y \frac{1}{y ^{3} z ^{3}} ((yz x)^{2} e^{yz x} - 2 (e^{yz x} yz x - e^{yz x}))

Simplify

2 y \frac{1}{y ^{3} z ^{3}} ((yz x)^{2} e^{yz x} - 2 (e^{yz x} yz x - e^{yz x})) : \frac{2 ( y ^{2} z ^{2} e ^{yz x} x ^{2} - 2 ( yz e ^{yz x} x - e ^{yz x} ) )}{y ^{2} z ^{3}}

= \frac{2 ( y ^{2} z ^{2} e ^{yz x} x ^{2} - 2 ( yz e ^{yz x} x - e ^{yz x} ) )}{y ^{2} z ^{3}}

Add a constant to the solution

= \frac{2 ( y ^{2} z ^{2} e ^{yz x} x ^{2} - 2 ( yz e ^{yz x} x - e ^{yz x} ) )}{y ^{2} z ^{3}} + C

\frac{\partial}{\partial x} (- 2 x z)

\frac{d}{d x} (\frac{8 x - 1}{3 x + 8})

f (x) = \frac{7 x}{x + 4}, (3, 3)

\int 7 x^{11} e^{- x^{6}} d x

\frac{\partial}{\partial x} (x y + yz)

What is the integral of 2x^2ye^{xyz} ?
The integral of 2x^2ye^{xyz} is (2(y^2z^2e^{yzx}x^2-2(yze^{yzx}x-e^{yzx})))/(y^2z^3)+C