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

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

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,

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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