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

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

Since the degree of the numerator is not less than the degree of the denominator, perform polynomial long division:

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

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

Simplify the integrand:

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

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

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

Rewrite the numerator of the integrand as $$$- c + 2 x=-2\left(c - x\right)+c$$$ and split the fraction:

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

Integrate term by term:

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

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

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

Let $$$u=c - x$$$.

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

Thus,

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

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

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

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

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

$$c \left(2 \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + \int{\frac{c}{\left(c - x\right)^{2}} d x}\right) + x = c \left(2 \ln{\left(\left|{{\color{red}{\left(c - x\right)}}}\right| \right)} + \int{\frac{c}{\left(c - x\right)^{2}} d x}\right) + x$$

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

$$c \left(2 \ln{\left(\left|{c - x}\right| \right)} + {\color{red}{\int{\frac{c}{\left(c - x\right)^{2}} d x}}}\right) + x = c \left(2 \ln{\left(\left|{c - x}\right| \right)} + {\color{red}{c \int{\frac{1}{\left(c - x\right)^{2}} d x}}}\right) + x$$

Let $$$u=c - x$$$.

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

The integral becomes

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

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

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

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

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

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

$$c \left(c {\color{red}{u}}^{-1} + 2 \ln{\left(\left|{c - x}\right| \right)}\right) + x = c \left(c {\color{red}{\left(c - x\right)}}^{-1} + 2 \ln{\left(\left|{c - x}\right| \right)}\right) + x$$

Therefore,

$$\int{\frac{x^{2}}{\left(c - x\right)^{2}} d x} = c \left(\frac{c}{c - x} + 2 \ln{\left(\left|{c - x}\right| \right)}\right) + x$$

Simplify:

$$\int{\frac{x^{2}}{\left(c - x\right)^{2}} d x} = \frac{c \left(c + 2 \left(c - x\right) \ln{\left(\left|{c - x}\right| \right)}\right) + x \left(c - x\right)}{c - x}$$

Add the constant of integration:

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

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