Integral of $$$x^{7} e^{- x^{8}}$$$

Find $$$\int x^{7} e^{- x^{8}}\, dx$$$.

Let $$$u=- x^{8}$$$.

Then $$$du=\left(- x^{8}\right)^{\prime }dx = - 8 x^{7} dx$$$ (steps can be seen »), and we have that $$$x^{7} dx = - \frac{du}{8}$$$.

So,

$${\color{red}{\int{x^{7} e^{- x^{8}} d x}}} = {\color{red}{\int{\left(- \frac{e^{u}}{8}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{8}$$$ and $$$f{\left(u \right)} = e^{u}$$$:

$${\color{red}{\int{\left(- \frac{e^{u}}{8}\right)d u}}} = {\color{red}{\left(- \frac{\int{e^{u} d u}}{8}\right)}}$$

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

$$- \frac{{\color{red}{\int{e^{u} d u}}}}{8} = - \frac{{\color{red}{e^{u}}}}{8}$$

Recall that $$$u=- x^{8}$$$:

$$- \frac{e^{{\color{red}{u}}}}{8} = - \frac{e^{{\color{red}{\left(- x^{8}\right)}}}}{8}$$

Therefore,

$$\int{x^{7} e^{- x^{8}} d x} = - \frac{e^{- x^{8}}}{8}$$

Add the constant of integration:

$$\int{x^{7} e^{- x^{8}} d x} = - \frac{e^{- x^{8}}}{8}+C$$

$$$\int x^{7} e^{- x^{8}}\, dx = - \frac{e^{- x^{8}}}{8} + C$$$A