What is the integral of-ze^{-yz} ?

The integral of-ze^{-yz} is e^{-zy}+C

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integral of-ze^{-yz}

\int - z e^{- yz} d y

e^{- zy} + C

Solution steps

\int - z e^{- yz} d y

Take the constant out:

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

= - z \cdot \int e^{- zy} d y

Apply u-substitution

: \int - \frac{e ^{u}}{z} d u

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

Take the constant out:

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

= - z (- \frac{1}{z} \cdot \int e^{u} d u)

Use the common integral:

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

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

Substitute back

u = - zy

= - z (- \frac{1}{z} e^{- zy})

Simplify

- z (- \frac{1}{z} e^{- zy}) : e^{- zy}

= e^{- zy}

Add a constant to the solution

= e^{- zy} + C

\int 10.2 + 2.5 x + 0.1 x^{2} d x

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\int \frac{x + 3}{x ^{2} + 2 x + 2} d x

x^{- 4}