Integral of $$$x e^{x^{2}}$$$

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

Let $$$u=x^{2}$$$.

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

The integral becomes

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

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

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

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

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

Recall that $$$u=x^{2}$$$:

$$\frac{e^{\color{red}{u}}}{2} = \frac{e^{\color{red}{x^{2}}}}{2}$$

Therefore,

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

Add the constant of integration:

$$\int{x e^{x^{2}} d x} = \frac{e^{x^{2}}}{2}+C$$

Answer: $$$\int{x e^{x^{2}} d x}=\frac{e^{x^{2}}}{2}+C$$$