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

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

Let $$$u=a - x$$$.

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

Thus,

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

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

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

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 $$$u=a - x$$$:

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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