Integral of $$$b^{2 x}$$$ with respect to $$$x$$$

Find $$$\int b^{2 x}\, dx$$$.

Let $$$u=2 x$$$.

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

So,

$${\color{red}{\int{b^{2 x} d x}}} = {\color{red}{\int{\frac{b^{u}}{2} d u}}}$$

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 $$$f{\left(u \right)} = b^{u}$$$:

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

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=b$$$:

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

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

$$\frac{b^{{\color{red}{u}}}}{2 \ln{\left(b \right)}} = \frac{b^{{\color{red}{\left(2 x\right)}}}}{2 \ln{\left(b \right)}}$$

Therefore,

$$\int{b^{2 x} d x} = \frac{b^{2 x}}{2 \ln{\left(b \right)}}$$

Add the constant of integration:

$$\int{b^{2 x} d x} = \frac{b^{2 x}}{2 \ln{\left(b \right)}}+C$$

$$$\int b^{2 x}\, dx = \frac{b^{2 x}}{2 \ln\left(b\right)} + C$$$A