Integral of $$$\left(2 x + 5\right)^{9}$$$

Find $$$\int \left(2 x + 5\right)^{9}\, dx$$$.

Let $$$u=2 x + 5$$$.

Then $$$du=\left(2 x + 5\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.

Therefore,

$${\color{red}{\int{\left(2 x + 5\right)^{9} d x}}} = {\color{red}{\int{\frac{u^{9}}{2} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = u^{9}$$$:

$${\color{red}{\int{\frac{u^{9}}{2} d u}}} = {\color{red}{\left(\frac{\int{u^{9} d u}}{2}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=9$$$:

$$\frac{{\color{red}{\int{u^{9} d u}}}}{2}=\frac{{\color{red}{\frac{u^{1 + 9}}{1 + 9}}}}{2}=\frac{{\color{red}{\left(\frac{u^{10}}{10}\right)}}}{2}$$

Recall that $$$u=2 x + 5$$$:

$$\frac{{\color{red}{u}}^{10}}{20} = \frac{{\color{red}{\left(2 x + 5\right)}}^{10}}{20}$$

Therefore,

$$\int{\left(2 x + 5\right)^{9} d x} = \frac{\left(2 x + 5\right)^{10}}{20}$$

Add the constant of integration:

$$\int{\left(2 x + 5\right)^{9} d x} = \frac{\left(2 x + 5\right)^{10}}{20}+C$$

$$$\int \left(2 x + 5\right)^{9}\, dx = \frac{\left(2 x + 5\right)^{10}}{20} + C$$$A