Integral of $$$3^{\sqrt{2} \sqrt{x}}$$$

Find $$$\int 3^{\sqrt{2} \sqrt{x}}\, dx$$$.

Change the base:

$${\color{red}{\int{3^{\sqrt{2} \sqrt{x}} d x}}} = {\color{red}{\int{e^{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}} d x}}}$$

Let $$$u=\sqrt{2} \sqrt{x} \ln{\left(3 \right)}$$$.

Then $$$du=\left(\sqrt{2} \sqrt{x} \ln{\left(3 \right)}\right)^{\prime }dx = \frac{\sqrt{2} \ln{\left(3 \right)}}{2 \sqrt{x}} dx$$$ (steps can be seen »), and we have that $$$\frac{dx}{\sqrt{x}} = \frac{\sqrt{2} du}{\ln{\left(3 \right)}}$$$.

The integral becomes

$${\color{red}{\int{e^{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}} d x}}} = {\color{red}{\int{\frac{u e^{u}}{\ln{\left(3 \right)}^{2}} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{\ln{\left(3 \right)}^{2}}$$$ and $$$f{\left(u \right)} = u e^{u}$$$:

$${\color{red}{\int{\frac{u e^{u}}{\ln{\left(3 \right)}^{2}} d u}}} = {\color{red}{\frac{\int{u e^{u} d u}}{\ln{\left(3 \right)}^{2}}}}$$

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

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

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

Therefore,

$$\frac{{\color{red}{\int{u e^{u} d u}}}}{\ln{\left(3 \right)}^{2}}=\frac{{\color{red}{\left(u \cdot e^{u}-\int{e^{u} \cdot 1 d u}\right)}}}{\ln{\left(3 \right)}^{2}}=\frac{{\color{red}{\left(u e^{u} - \int{e^{u} d u}\right)}}}{\ln{\left(3 \right)}^{2}}$$

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

$$\frac{u e^{u} - {\color{red}{\int{e^{u} d u}}}}{\ln{\left(3 \right)}^{2}} = \frac{u e^{u} - {\color{red}{e^{u}}}}{\ln{\left(3 \right)}^{2}}$$

Recall that $$$u=\sqrt{2} \sqrt{x} \ln{\left(3 \right)}$$$:

$$\frac{- e^{{\color{red}{u}}} + {\color{red}{u}} e^{{\color{red}{u}}}}{\ln{\left(3 \right)}^{2}} = \frac{- e^{{\color{red}{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}}}} + {\color{red}{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}}} e^{{\color{red}{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}}}}}{\ln{\left(3 \right)}^{2}}$$

Therefore,

$$\int{3^{\sqrt{2} \sqrt{x}} d x} = \frac{\sqrt{2} \sqrt{x} e^{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}} \ln{\left(3 \right)} - e^{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}}}{\ln{\left(3 \right)}^{2}}$$

Simplify:

$$\int{3^{\sqrt{2} \sqrt{x}} d x} = \frac{\left(\sqrt{2} \sqrt{x} \ln{\left(3 \right)} - 1\right) e^{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}}}{\ln{\left(3 \right)}^{2}}$$

Add the constant of integration:

$$\int{3^{\sqrt{2} \sqrt{x}} d x} = \frac{\left(\sqrt{2} \sqrt{x} \ln{\left(3 \right)} - 1\right) e^{\sqrt{2} \sqrt{x} \ln{\left(3 \right)}}}{\ln{\left(3 \right)}^{2}}+C$$

$$$\int 3^{\sqrt{2} \sqrt{x}}\, dx = \frac{\left(\sqrt{2} \sqrt{x} \ln\left(3\right) - 1\right) e^{\sqrt{2} \sqrt{x} \ln\left(3\right)}}{\ln^{2}\left(3\right)} + C$$$A