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

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

Let $$$u=1 - x$$$.

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

Therefore,

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

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

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-3$$$:

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

Recall that $$$u=1 - x$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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