What is the integral of x^3ze^{xz} ?

The integral of x^3ze^{xz} is e^{zx}x^3-(3(z^2e^{zx}x^2-2(ze^{zx}x-e^{zx})))/(z^3)+C

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integral of x^3ze^{xz}

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{e ^{u} u ^{3}}{z ^{4}} d u

= z \cdot \int \frac{e ^{u} u ^{3}}{z ^{4}} d u

Take the constant out:

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

= z \frac{1}{z ^{4}} \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

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

= z \frac{1}{z ^{4}} (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}))

= z \frac{1}{z ^{4}} (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = z x

= z \frac{1}{z ^{4}} ((z x)^{3} e^{z x} - 3 ((z x)^{2} e^{z x} - 2 (e^{z x} z x - e^{z x})))

Simplify

z \frac{1}{z ^{4}} ((z x)^{3} e^{z x} - 3 ((z x)^{2} e^{z x} - 2 (e^{z x} z x - e^{z x}))) : e^{z x} x^{3} - \frac{3 ( z ^{2} e ^{z x} x ^{2} - 2 ( z e ^{z x} x - e ^{z x} ) )}{z ^{3}}

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

Add a constant to the solution

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

n = 0 \sum \infty - 6 (- \frac{1}{3})^{2 n}

f (x) = 2 x^{3} ln (x)

\frac{d}{d x} ((2 x^{2} - 4)^{3})

\int sin (3 x) sin (n x) d x

(1 - t) (3 + t^{2})^{- 1}

What is the integral of x^3ze^{xz} ?
The integral of x^3ze^{xz} is e^{zx}x^3-(3(z^2e^{zx}x^2-2(ze^{zx}x-e^{zx})))/(z^3)+C