What is the integral of e^{z/y} ?

The integral of e^{z/y} is ye^{z/y}+C

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integral of e^{z/y}

\int e^{\frac{z}{y}} d z

y e^{\frac{z}{y}} + C

Solution steps

\int e^{\frac{z}{y}} d z

Apply u-substitution

: \int e^{u} y d u

= \int e^{u} y d u

Take the constant out:

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

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

Use the common integral:

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

= y e^{u}

Substitute back

u = \frac{z}{y}

= y e^{\frac{z}{y}}

Add a constant to the solution

= y e^{\frac{z}{y}} + C

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

\frac{\partial}{\partial x} (1 + x^{2} y - x y)

\frac{d}{d t} (2 t + 2)

\frac{d}{d x} (3 (x + 1)^{2})

\frac{4 s}{s ^{2} - 16}