Integral of $$$e^{\frac{u}{v}}$$$ with respect to $$$u$$$

Find $$$\int e^{\frac{u}{v}}\, du$$$.

Let $$$w=\frac{u}{v}$$$.

Then $$$dw=\left(\frac{u}{v}\right)^{\prime }du = \frac{du}{v}$$$ (steps can be seen »), and we have that $$$du = v dw$$$.

The integral becomes

$${\color{red}{\int{e^{\frac{u}{v}} d u}}} = {\color{red}{\int{v e^{w} d w}}}$$

Apply the constant multiple rule $$$\int c f{\left(w \right)}\, dw = c \int f{\left(w \right)}\, dw$$$ with $$$c=v$$$ and $$$f{\left(w \right)} = e^{w}$$$:

$${\color{red}{\int{v e^{w} d w}}} = {\color{red}{v \int{e^{w} d w}}}$$

The integral of the exponential function is $$$\int{e^{w} d w} = e^{w}$$$:

$$v {\color{red}{\int{e^{w} d w}}} = v {\color{red}{e^{w}}}$$

Recall that $$$w=\frac{u}{v}$$$:

$$v e^{{\color{red}{w}}} = v e^{{\color{red}{\frac{u}{v}}}}$$

Therefore,

$$\int{e^{\frac{u}{v}} d u} = v e^{\frac{u}{v}}$$

Add the constant of integration:

$$\int{e^{\frac{u}{v}} d u} = v e^{\frac{u}{v}}+C$$

$$$\int e^{\frac{u}{v}}\, du = v e^{\frac{u}{v}} + C$$$A