Solve A ∫ f(x)=x^2-3x+2

$$A \int f ( x ) = x ^ { 2 } - 3 x + 2$$

Solve for A

$$\left\{\begin{matrix}A=-\frac{2\left(1-x\right)\left(x-2\right)}{fx^{2}+2С}\text{, }&С\neq -\frac{fx^{2}}{2}\\A\in \mathrm{R}\text{, }&\left(x=2\text{ and }f=-\frac{С}{2}\right)\text{ or }\left(x=1\text{ and }f=-2С_{1}\right)\end{matrix}\right.$$

Steps for Solving Linear Equation

The equation is in standard form.

$$\left(\frac{fx^{2}}{2}+С\right)A=x^{2}-3x+2$$

Divide both sides by \frac{1}{2}fx^{2}+С.

$$\frac{\left(\frac{fx^{2}}{2}+С\right)A}{\frac{fx^{2}}{2}+С}=\frac{\left(x-2\right)\left(x-1\right)}{\frac{fx^{2}}{2}+С}$$

Dividing by \frac{1}{2}fx^{2}+С undoes the multiplication by \frac{1}{2}fx^{2}+С.

$$A=\frac{\left(x-2\right)\left(x-1\right)}{\frac{fx^{2}}{2}+С}$$

Divide \left(-2+x\right)\left(-1+x\right) by \frac{1}{2}fx^{2}+С.

$$A=\frac{2\left(x-2\right)\left(x-1\right)}{fx^{2}+2С}$$

Solve for f

$$\left\{\begin{matrix}f=-\frac{2С}{x^{2}}+\frac{2\left(x^{2}-3x+2\right)}{Ax^{2}}\text{, }&A\neq 0\text{ and }x\neq 0\\f\in \mathrm{R}\text{, }&\left(A=\frac{2}{С}\text{ and }x=0\text{ and }С_{1}\neq 0\right)\text{ or }\left(A=0\text{ and }x=2\right)\text{ or }\left(A=0\text{ and }x=1\right)\end{matrix}\right.$$