Integral of $$$e^{- \sqrt{x}}$$$

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

Let $$$u=- \sqrt{x}$$$.

Then $$$du=\left(- \sqrt{x}\right)^{\prime }dx = - \frac{1}{2 \sqrt{x}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{\sqrt{x}} = - 2 du$$$.

Therefore,

$${\color{red}{\int{e^{- \sqrt{x}} d x}}} = {\color{red}{\int{2 u e^{u} d u}}}$$

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

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

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

Let $$$\operatorname{m}=u$$$ and $$$\operatorname{dv}=e^{u} du$$$.

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

So,

$$2 {\color{red}{\int{u e^{u} d u}}}=2 {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}=2 {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}$$

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

$$2 u e^{u} - 2 {\color{red}{\int{e^{u} d u}}} = 2 u e^{u} - 2 {\color{red}{e^{u}}}$$

Recall that $$$u=- \sqrt{x}$$$:

$$- 2 e^{{\color{red}{u}}} + 2 {\color{red}{u}} e^{{\color{red}{u}}} = - 2 e^{{\color{red}{\left(- \sqrt{x}\right)}}} + 2 {\color{red}{\left(- \sqrt{x}\right)}} e^{{\color{red}{\left(- \sqrt{x}\right)}}}$$

Therefore,

$$\int{e^{- \sqrt{x}} d x} = - 2 \sqrt{x} e^{- \sqrt{x}} - 2 e^{- \sqrt{x}}$$

Simplify:

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

Add the constant of integration:

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

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