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

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

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

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

Thus,

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

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 $$$u=x - 1$$$:

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

Therefore,

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

Add the constant of integration:

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

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