Integral of $$$e^{\sqrt{33} \sqrt{x}}$$$

Find $$$\int e^{\sqrt{33} \sqrt{x}}\, dx$$$.

Let $$$u=\sqrt{33} \sqrt{x}$$$.

Then $$$du=\left(\sqrt{33} \sqrt{x}\right)^{\prime }dx = \frac{\sqrt{33}}{2 \sqrt{x}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{\sqrt{x}} = \frac{2 \sqrt{33} du}{33}$$$.

The integral becomes

$${\color{red}{\int{e^{\sqrt{33} \sqrt{x}} d x}}} = {\color{red}{\int{\frac{2 u e^{u}}{33} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{2}{33}$$$ and $$$f{\left(u \right)} = u e^{u}$$$:

$${\color{red}{\int{\frac{2 u e^{u}}{33} d u}}} = {\color{red}{\left(\frac{2 \int{u e^{u} d u}}{33}\right)}}$$

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

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

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

Thus,

$$\frac{2 {\color{red}{\int{u e^{u} d u}}}}{33}=\frac{2 {\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}}{33}=\frac{2 {\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}}{33}$$

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

$$\frac{2 u e^{u}}{33} - \frac{2 {\color{red}{\int{e^{u} d u}}}}{33} = \frac{2 u e^{u}}{33} - \frac{2 {\color{red}{e^{u}}}}{33}$$

Recall that $$$u=\sqrt{33} \sqrt{x}$$$:

$$- \frac{2 e^{{\color{red}{u}}}}{33} + \frac{2 {\color{red}{u}} e^{{\color{red}{u}}}}{33} = - \frac{2 e^{{\color{red}{\sqrt{33} \sqrt{x}}}}}{33} + \frac{2 {\color{red}{\sqrt{33} \sqrt{x}}} e^{{\color{red}{\sqrt{33} \sqrt{x}}}}}{33}$$

Therefore,

$$\int{e^{\sqrt{33} \sqrt{x}} d x} = \frac{2 \sqrt{33} \sqrt{x} e^{\sqrt{33} \sqrt{x}}}{33} - \frac{2 e^{\sqrt{33} \sqrt{x}}}{33}$$

Simplify:

$$\int{e^{\sqrt{33} \sqrt{x}} d x} = \frac{2 \left(\sqrt{33} \sqrt{x} - 1\right) e^{\sqrt{33} \sqrt{x}}}{33}$$

Add the constant of integration:

$$\int{e^{\sqrt{33} \sqrt{x}} d x} = \frac{2 \left(\sqrt{33} \sqrt{x} - 1\right) e^{\sqrt{33} \sqrt{x}}}{33}+C$$

$$$\int e^{\sqrt{33} \sqrt{x}}\, dx = \frac{2 \left(\sqrt{33} \sqrt{x} - 1\right) e^{\sqrt{33} \sqrt{x}}}{33} + C$$$A