Integral of $$$\frac{e^{\frac{1}{x^{3}}}}{x^{4}}$$$

Find $$$\int \frac{e^{\frac{1}{x^{3}}}}{x^{4}}\, dx$$$.

Let $$$u=x^{3}$$$.

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

The integral can be rewritten as

$${\color{red}{\int{\frac{e^{\frac{1}{x^{3}}}}{x^{4}} d x}}} = {\color{red}{\int{\frac{e^{\frac{1}{u}}}{3 u^{2}} d u}}}$$

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

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

Let $$$v=\frac{1}{u}$$$.

Then $$$dv=\left(\frac{1}{u}\right)^{\prime }du = - \frac{1}{u^{2}} du$$$ (steps can be seen »), and we have that $$$\frac{du}{u^{2}} = - dv$$$.

So,

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

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

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

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

$$- \frac{{\color{red}{\int{e^{v} d v}}}}{3} = - \frac{{\color{red}{e^{v}}}}{3}$$

Recall that $$$v=\frac{1}{u}$$$:

$$- \frac{e^{{\color{red}{v}}}}{3} = - \frac{e^{{\color{red}{\frac{1}{u}}}}}{3}$$

Recall that $$$u=x^{3}$$$:

$$- \frac{e^{{\color{red}{u}}^{-1}}}{3} = - \frac{e^{{\color{red}{x^{3}}}^{-1}}}{3}$$

Therefore,

$$\int{\frac{e^{\frac{1}{x^{3}}}}{x^{4}} d x} = - \frac{e^{\frac{1}{x^{3}}}}{3}$$

Add the constant of integration:

$$\int{\frac{e^{\frac{1}{x^{3}}}}{x^{4}} d x} = - \frac{e^{\frac{1}{x^{3}}}}{3}+C$$

$$$\int \frac{e^{\frac{1}{x^{3}}}}{x^{4}}\, dx = - \frac{e^{\frac{1}{x^{3}}}}{3} + C$$$A