Solve ∫ (2x^2-5x+3) wrt x=2x^3/3-5x^2/2+3x+c

$$\int ( 2 x ^ { 2 } - 5 x + 3 ) d x = \frac { 2 x ^ { 3 } } { 3 } - \frac { 5 x ^ { 2 } } { 2 } + 3 x + c$$

Solve for c

$$c=С$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$6\int 2x^{2}-5x+3\mathrm{d}x=2\times 2x^{3}-3\times 5x^{2}+18x+6c$$

Do the multiplications.

$$6\int 2x^{2}-5x+3\mathrm{d}x=4x^{3}-15x^{2}+18x+6c$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{3}-15x^{2}+18x+6c=6\int 2x^{2}-5x+3\mathrm{d}x$$

Subtract 4x^{3} from both sides.

$$-15x^{2}+18x+6c=6\int 2x^{2}-5x+3\mathrm{d}x-4x^{3}$$

Add 15x^{2} to both sides.

$$18x+6c=6\int 2x^{2}-5x+3\mathrm{d}x-4x^{3}+15x^{2}$$

Subtract 18x from both sides.

$$6c=6\int 2x^{2}-5x+3\mathrm{d}x-4x^{3}+15x^{2}-18x$$

The equation is in standard form.

$$6c=С$$

Divide both sides by 6.

$$\frac{6c}{6}=\frac{С}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$c=\frac{С}{6}$$

Solve for x

$$x\in \mathrm{R}$$

$$c=С$$