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 »).

The integral can be rewritten as

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

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 $$$f{\left(x \right)} = 2^{- x}$$$:

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

Let $$$u=- x$$$.

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

Therefore,

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

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

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

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

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

Recall that $$$u=- x$$$:

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

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