Integral of $$$20 e^{\frac{3 x}{2}}$$$

Find $$$\int 20 e^{\frac{3 x}{2}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=20$$$ and $$$f{\left(x \right)} = e^{\frac{3 x}{2}}$$$:

$${\color{red}{\int{20 e^{\frac{3 x}{2}} d x}}} = {\color{red}{\left(20 \int{e^{\frac{3 x}{2}} d x}\right)}}$$

Let $$$u=\frac{3 x}{2}$$$.

Then $$$du=\left(\frac{3 x}{2}\right)^{\prime }dx = \frac{3 dx}{2}$$$ (steps can be seen »), and we have that $$$dx = \frac{2 du}{3}$$$.

The integral becomes

$$20 {\color{red}{\int{e^{\frac{3 x}{2}} d x}}} = 20 {\color{red}{\int{\frac{2 e^{u}}{3} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{2}{3}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$20 {\color{red}{\int{\frac{2 e^{u}}{3} d u}}} = 20 {\color{red}{\left(\frac{2 \int{e^{u} d u}}{3}\right)}}$$

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

$$\frac{40 {\color{red}{\int{e^{u} d u}}}}{3} = \frac{40 {\color{red}{e^{u}}}}{3}$$

Recall that $$$u=\frac{3 x}{2}$$$:

$$\frac{40 e^{{\color{red}{u}}}}{3} = \frac{40 e^{{\color{red}{\left(\frac{3 x}{2}\right)}}}}{3}$$

Therefore,

$$\int{20 e^{\frac{3 x}{2}} d x} = \frac{40 e^{\frac{3 x}{2}}}{3}$$

Add the constant of integration:

$$\int{20 e^{\frac{3 x}{2}} d x} = \frac{40 e^{\frac{3 x}{2}}}{3}+C$$

$$$\int 20 e^{\frac{3 x}{2}}\, dx = \frac{40 e^{\frac{3 x}{2}}}{3} + C$$$A