Integral of $$$e^{\frac{u}{2}}$$$

Find $$$\int e^{\frac{u}{2}}\, du$$$.

Let $$$v=\frac{u}{2}$$$.

Then $$$dv=\left(\frac{u}{2}\right)^{\prime }du = \frac{du}{2}$$$ (steps can be seen »), and we have that $$$du = 2 dv$$$.

Thus,

$${\color{red}{\int{e^{\frac{u}{2}} d u}}} = {\color{red}{\int{2 e^{v} d v}}}$$

Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=2$$$ and $$$f{\left(v \right)} = e^{v}$$$:

$${\color{red}{\int{2 e^{v} d v}}} = {\color{red}{\left(2 \int{e^{v} d v}\right)}}$$

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

$$2 {\color{red}{\int{e^{v} d v}}} = 2 {\color{red}{e^{v}}}$$

Recall that $$$v=\frac{u}{2}$$$:

$$2 e^{{\color{red}{v}}} = 2 e^{{\color{red}{\left(\frac{u}{2}\right)}}}$$

Therefore,

$$\int{e^{\frac{u}{2}} d u} = 2 e^{\frac{u}{2}}$$

Add the constant of integration:

$$\int{e^{\frac{u}{2}} d u} = 2 e^{\frac{u}{2}}+C$$

$$$\int e^{\frac{u}{2}}\, du = 2 e^{\frac{u}{2}} + C$$$A