Integral of $$$- \frac{6}{\left(x - 1\right)^{2}}$$$

Find $$$\int \left(- \frac{6}{\left(x - 1\right)^{2}}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-6$$$ and $$$f{\left(x \right)} = \frac{1}{\left(x - 1\right)^{2}}$$$:

$${\color{red}{\int{\left(- \frac{6}{\left(x - 1\right)^{2}}\right)d x}}} = {\color{red}{\left(- 6 \int{\frac{1}{\left(x - 1\right)^{2}} d x}\right)}}$$

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,

$$- 6 {\color{red}{\int{\frac{1}{\left(x - 1\right)^{2}} d x}}} = - 6 {\color{red}{\int{\frac{1}{u^{2}} d u}}}$$

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

$$- 6 {\color{red}{\int{\frac{1}{u^{2}} d u}}}=- 6 {\color{red}{\int{u^{-2} d u}}}=- 6 {\color{red}{\frac{u^{-2 + 1}}{-2 + 1}}}=- 6 {\color{red}{\left(- u^{-1}\right)}}=- 6 {\color{red}{\left(- \frac{1}{u}\right)}}$$

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

$$6 {\color{red}{u}}^{-1} = 6 {\color{red}{\left(x - 1\right)}}^{-1}$$

Therefore,

$$\int{\left(- \frac{6}{\left(x - 1\right)^{2}}\right)d x} = \frac{6}{x - 1}$$

Add the constant of integration:

$$\int{\left(- \frac{6}{\left(x - 1\right)^{2}}\right)d x} = \frac{6}{x - 1}+C$$

$$$\int \left(- \frac{6}{\left(x - 1\right)^{2}}\right)\, dx = \frac{6}{x - 1} + C$$$A