What is the integral of e^{y(1-x)} ?

The integral of e^{y(1-x)} is -1/y e^{-y(-1+x)}+C

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integral of e^{y(1-x)}

\int e^{y (1 - x)} d x

- \frac{1}{y} e^{- y (- 1 + x)} + C

Solution steps

\int e^{y (1 - x)} d x

Factor

1 - x : - (- 1 + x)

= \int e^{(- (- 1 + x)) y} d x

Simplify

= \int e^{- y (- 1 + x)} d x

Apply u-substitution

= \int - \frac{e ^{u}}{y} d u

Take the constant out:

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

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

Use the common integral:

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

= - \frac{1}{y} e^{u}

Substitute back

u = - y (- 1 + x)

= - \frac{1}{y} e^{- y (- 1 + x)}

Add a constant to the solution

= - \frac{1}{y} e^{- y (- 1 + x)} + C

What is the integral of e^{y(1-x)} ?
The integral of e^{y(1-x)} is -1/y e^{-y(-1+x)}+C