Integral of $x e^{- x}$

Find $\int x e^{- x}\, dx$.

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

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

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

The integral becomes

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

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

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

Let $u=- x$.

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

Thus,

$$- x e^{- x} + {\color{red}{\int{e^{- x} d x}}} = - x e^{- x} + {\color{red}{\int{\left(- e^{u}\right)d u}}}$$

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

$$- x e^{- x} + {\color{red}{\int{\left(- e^{u}\right)d u}}} = - x e^{- x} + {\color{red}{\left(- \int{e^{u} d u}\right)}}$$

The integral of the exponential function is $\int{e^{u} d u} = e^{u}$:

$$- x e^{- x} - {\color{red}{\int{e^{u} d u}}} = - x e^{- x} - {\color{red}{e^{u}}}$$

Recall that $u=- x$:

$$- x e^{- x} - e^{{\color{red}{u}}} = - x e^{- x} - e^{{\color{red}{\left(- x\right)}}}$$

Therefore,

$$\int{x e^{- x} d x} = - x e^{- x} - e^{- x}$$

Simplify:

$$\int{x e^{- x} d x} = \left(- x - 1\right) e^{- x}$$

Add the constant of integration:

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

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