Integral of $$$\frac{x^{3}}{\left(1 - x^{4}\right)^{2}}$$$

Find $$$\int \frac{x^{3}}{\left(1 - x^{4}\right)^{2}}\, dx$$$.

Let $$$u=1 - x^{4}$$$.

Then $$$du=\left(1 - x^{4}\right)^{\prime }dx = - 4 x^{3} dx$$$ (steps can be seen »), and we have that $$$x^{3} dx = - \frac{du}{4}$$$.

The integral becomes

$${\color{red}{\int{\frac{x^{3}}{\left(1 - x^{4}\right)^{2}} d x}}} = {\color{red}{\int{\left(- \frac{1}{4 u^{2}}\right)d u}}}$$

Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=- \frac{1}{4}$$$ and $$$f{\left(u \right)} = \frac{1}{u^{2}}$$$:

$${\color{red}{\int{\left(- \frac{1}{4 u^{2}}\right)d u}}} = {\color{red}{\left(- \frac{\int{\frac{1}{u^{2}} d u}}{4}\right)}}$$

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-2$$$:

$$- \frac{{\color{red}{\int{\frac{1}{u^{2}} d u}}}}{4}=- \frac{{\color{red}{\int{u^{-2} d u}}}}{4}=- \frac{{\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}}{4}=- \frac{{\color{red}{\left(- u^{-1}\right)}}}{4}=- \frac{{\color{red}{\left(- \frac{1}{u}\right)}}}{4}$$

Recall that $$$u=1 - x^{4}$$$:

$$\frac{{\color{red}{u}}^{-1}}{4} = \frac{{\color{red}{\left(1 - x^{4}\right)}}^{-1}}{4}$$

Therefore,

$$\int{\frac{x^{3}}{\left(1 - x^{4}\right)^{2}} d x} = \frac{1}{4 \left(1 - x^{4}\right)}$$

Simplify:

$$\int{\frac{x^{3}}{\left(1 - x^{4}\right)^{2}} d x} = - \frac{1}{4 x^{4} - 4}$$

Add the constant of integration:

$$\int{\frac{x^{3}}{\left(1 - x^{4}\right)^{2}} d x} = - \frac{1}{4 x^{4} - 4}+C$$

$$$\int \frac{x^{3}}{\left(1 - x^{4}\right)^{2}}\, dx = - \frac{1}{4 x^{4} - 4} + C$$$A