Integral of $$$\sqrt{3 x + 6}$$$

Find $$$\int \sqrt{3 x + 6}\, dx$$$.

Let $$$u=3 x + 6$$$.

Then $$$du=\left(3 x + 6\right)^{\prime }dx = 3 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{3}$$$.

So,

$${\color{red}{\int{\sqrt{3 x + 6} d x}}} = {\color{red}{\int{\frac{\sqrt{u}}{3} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(u \right)} = \sqrt{u}$$$:

$${\color{red}{\int{\frac{\sqrt{u}}{3} d u}}} = {\color{red}{\left(\frac{\int{\sqrt{u} d u}}{3}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=\frac{1}{2}$$$:

$$\frac{{\color{red}{\int{\sqrt{u} d u}}}}{3}=\frac{{\color{red}{\int{u^{\frac{1}{2}} d u}}}}{3}=\frac{{\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{3}=\frac{{\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}}{3}$$

Recall that $$$u=3 x + 6$$$:

$$\frac{2 {\color{red}{u}}^{\frac{3}{2}}}{9} = \frac{2 {\color{red}{\left(3 x + 6\right)}}^{\frac{3}{2}}}{9}$$

Therefore,

$$\int{\sqrt{3 x + 6} d x} = \frac{2 \left(3 x + 6\right)^{\frac{3}{2}}}{9}$$

Simplify:

$$\int{\sqrt{3 x + 6} d x} = \frac{2 \sqrt{3} \left(x + 2\right)^{\frac{3}{2}}}{3}$$

Add the constant of integration:

$$\int{\sqrt{3 x + 6} d x} = \frac{2 \sqrt{3} \left(x + 2\right)^{\frac{3}{2}}}{3}+C$$

$$$\int \sqrt{3 x + 6}\, dx = \frac{2 \sqrt{3} \left(x + 2\right)^{\frac{3}{2}}}{3} + C$$$A