Integral of $$$2^{x} x$$$

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

For the integral $$$\int{2^{x} x d x}$$$, use integration by parts $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Let $$$\operatorname{u}=x$$$ and $$$\operatorname{dv}=2^{x} dx$$$.

Then $$$\operatorname{du}=\left(x\right)^{\prime }dx=1 dx$$$ (steps can be seen ») and $$$\operatorname{v}=\int{2^{x} d x}=\frac{2^{x}}{\ln{\left(2 \right)}}$$$ (steps can be seen »).

So,

$${\color{red}{\int{2^{x} x d x}}}={\color{red}{\left(x \cdot \frac{2^{x}}{\ln{\left(2 \right)}}-\int{\frac{2^{x}}{\ln{\left(2 \right)}} \cdot 1 d x}\right)}}={\color{red}{\left(\frac{2^{x} x}{\ln{\left(2 \right)}} - \int{\frac{2^{x}}{\ln{\left(2 \right)}} d x}\right)}}$$

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

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

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=2$$$:

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

Therefore,

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

Simplify:

$$\int{2^{x} x d x} = \frac{2^{x} \left(x \ln{\left(2 \right)} - 1\right)}{\ln{\left(2 \right)}^{2}}$$

Add the constant of integration:

$$\int{2^{x} x d x} = \frac{2^{x} \left(x \ln{\left(2 \right)} - 1\right)}{\ln{\left(2 \right)}^{2}}+C$$

$$$\int 2^{x} x\, dx = \frac{2^{x} \left(x \ln\left(2\right) - 1\right)}{\ln^{2}\left(2\right)} + C$$$A