Integral of $$$f x \left(x - 1\right)$$$ with respect to $$$x$$$

Find $$$\int f x \left(x - 1\right)\, dx$$$.

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

$${\color{red}{\int{f x \left(x - 1\right) d x}}} = {\color{red}{f \int{x \left(x - 1\right) d x}}}$$

Expand the expression:

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

Integrate term by term:

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

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

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

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

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

Therefore,

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

Simplify:

$$\int{f x \left(x - 1\right) d x} = \frac{f x^{2} \left(2 x - 3\right)}{6}$$

Add the constant of integration:

$$\int{f x \left(x - 1\right) d x} = \frac{f x^{2} \left(2 x - 3\right)}{6}+C$$

$$$\int f x \left(x - 1\right)\, dx = \frac{f x^{2} \left(2 x - 3\right)}{6} + C$$$A