Solve ∫ (3-2x-x^4) wrt x=3x-x^2-x^5/4+c

$$\int ( 3 - 2 x - x ^ { 4 } ) d x = 3 x - x ^ { 2 } - \frac { x ^ { 5 } } { 4 } + c$$

Solve for c

$$c=\frac{x^{5}}{20}-С$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

$$4\int 3-2x-x^{4}\mathrm{d}x=12x-4x^{2}-x^{5}+4c$$

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

$$12x-4x^{2}-x^{5}+4c=4\int 3-2x-x^{4}\mathrm{d}x$$

Subtract 12x from both sides.

$$-4x^{2}-x^{5}+4c=4\int 3-2x-x^{4}\mathrm{d}x-12x$$

Add 4x^{2} to both sides.

$$-x^{5}+4c=4\int 3-2x-x^{4}\mathrm{d}x-12x+4x^{2}$$

Add x^{5} to both sides.

$$4c=4\int 3-2x-x^{4}\mathrm{d}x-12x+4x^{2}+x^{5}$$

The equation is in standard form.

$$4c=\frac{x^{5}}{5}+4С$$

Divide both sides by 4.

$$\frac{4c}{4}=\frac{\frac{x^{5}}{5}+4С}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$c=\frac{\frac{x^{5}}{5}+4С}{4}$$

Divide \frac{x^{5}}{5}+4С by 4.

$$c=\frac{x^{5}}{20}+С$$

Solve for x

$$x=\sqrt[5]{С+20c}$$