What is the integral of e^{xyz} ?

The integral of e^{xyz} is 1/(yz)e^{yzx}+C

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integral of e^{xyz}

\int e^{x yz} d x

\frac{1}{yz} e^{yz x} + C

Solution steps

\int e^{x yz} d x

Apply u-substitution

: \int \frac{e ^{u}}{yz} d u

= \int \frac{e ^{u}}{yz} d u

Take the constant out:

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

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

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{yz} e^{u}

Substitute back

u = yz x

= \frac{1}{yz} e^{yz x}

Add a constant to the solution

= \frac{1}{yz} e^{yz x} + C

x \to 1 lim (sin (3 π x))

y + (x + 1) \frac{d y}{d x} = 2 x + 2, y (0) = 2

y^{^{'}} = - 8 x y

\frac{d y}{d x} = e^{2 x - y}

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