Integral of $$$x^{3} e^{x^{4}}$$$

Find $$$\int x^{3} e^{x^{4}}\, dx$$$.

Let $$$u=x^{4}$$$.

Then $$$du=\left(x^{4}\right)^{\prime }dx = 4 x^{3} dx$$$ (steps can be seen »), and we have that $$$x^{3} dx = \frac{du}{4}$$$.

Thus,

$${\color{red}{\int{x^{3} e^{x^{4}} d x}}} = {\color{red}{\int{\frac{e^{u}}{4} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{4}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{e^{u}}{4} d u}}} = {\color{red}{\left(\frac{\int{e^{u} d u}}{4}\right)}}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{4} = \frac{{\color{red}{e^{u}}}}{4}$$

Recall that $$$u=x^{4}$$$:

$$\frac{e^{{\color{red}{u}}}}{4} = \frac{e^{{\color{red}{x^{4}}}}}{4}$$

Therefore,

$$\int{x^{3} e^{x^{4}} d x} = \frac{e^{x^{4}}}{4}$$

Add the constant of integration:

$$\int{x^{3} e^{x^{4}} d x} = \frac{e^{x^{4}}}{4}+C$$

$$$\int x^{3} e^{x^{4}}\, dx = \frac{e^{x^{4}}}{4} + C$$$A