Integral of $$$15 \sqrt{5} e^{5 x}$$$

Find $$$\int 15 \sqrt{5} e^{5 x}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=15 \sqrt{5}$$$ and $$$f{\left(x \right)} = e^{5 x}$$$:

$${\color{red}{\int{15 \sqrt{5} e^{5 x} d x}}} = {\color{red}{\left(15 \sqrt{5} \int{e^{5 x} d x}\right)}}$$

Let $$$u=5 x$$$.

Then $$$du=\left(5 x\right)^{\prime }dx = 5 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{5}$$$.

So,

$$15 \sqrt{5} {\color{red}{\int{e^{5 x} d x}}} = 15 \sqrt{5} {\color{red}{\int{\frac{e^{u}}{5} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{5}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$$15 \sqrt{5} {\color{red}{\int{\frac{e^{u}}{5} d u}}} = 15 \sqrt{5} {\color{red}{\left(\frac{\int{e^{u} d u}}{5}\right)}}$$

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

$$3 \sqrt{5} {\color{red}{\int{e^{u} d u}}} = 3 \sqrt{5} {\color{red}{e^{u}}}$$

Recall that $$$u=5 x$$$:

$$3 \sqrt{5} e^{{\color{red}{u}}} = 3 \sqrt{5} e^{{\color{red}{\left(5 x\right)}}}$$

Therefore,

$$\int{15 \sqrt{5} e^{5 x} d x} = 3 \sqrt{5} e^{5 x}$$

Add the constant of integration:

$$\int{15 \sqrt{5} e^{5 x} d x} = 3 \sqrt{5} e^{5 x}+C$$

$$$\int 15 \sqrt{5} e^{5 x}\, dx = 3 \sqrt{5} e^{5 x} + C$$$A