Integral of $$$x^{6} e^{x^{7}}$$$

Find $$$\int x^{6} e^{x^{7}}\, dx$$$.

Let $$$u=x^{7}$$$.

Then $$$du=\left(x^{7}\right)^{\prime }dx = 7 x^{6} dx$$$ (steps can be seen »), and we have that $$$x^{6} dx = \frac{du}{7}$$$.

The integral can be rewritten as

$${\color{red}{\int{x^{6} e^{x^{7}} d x}}} = {\color{red}{\int{\frac{e^{u}}{7} d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{7}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\frac{e^{u}}{7} d u}}} = {\color{red}{\left(\frac{\int{e^{u} d u}}{7}\right)}}$$

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

$$\frac{{\color{red}{\int{e^{u} d u}}}}{7} = \frac{{\color{red}{e^{u}}}}{7}$$

Recall that $$$u=x^{7}$$$:

$$\frac{e^{{\color{red}{u}}}}{7} = \frac{e^{{\color{red}{x^{7}}}}}{7}$$

Therefore,

$$\int{x^{6} e^{x^{7}} d x} = \frac{e^{x^{7}}}{7}$$

Add the constant of integration:

$$\int{x^{6} e^{x^{7}} d x} = \frac{e^{x^{7}}}{7}+C$$

$$$\int x^{6} e^{x^{7}}\, dx = \frac{e^{x^{7}}}{7} + C$$$A