Solve wrt y = ∫ (2x^3+3x-2x^3)dx

$$d y = \int ( 2 x ^ { 3 } + 3 x - 2 x ^ { 3 } ) d x$$

Solve for d

$$\left\{\begin{matrix}d=\frac{3x^{2}}{2y}+\frac{С}{y}\text{, }&y\neq 0\\d\in \mathrm{R}\text{, }&С=\frac{3x^{2}}{2}\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine 2x^{3} and -2x^{3} to get 0.

$$dy=\int 3x\mathrm{d}x$$

The equation is in standard form.

$$yd=\frac{3x^{2}}{2}+С$$

Divide both sides by y.

$$\frac{yd}{y}=\frac{\frac{3x^{2}}{2}+С}{y}$$

Dividing by y undoes the multiplication by y.

$$d=\frac{\frac{3x^{2}}{2}+С}{y}$$

Solve for y

$$\left\{\begin{matrix}y=\frac{3x^{2}}{2d}+\frac{С}{d}\text{, }&d\neq 0\\y\in \mathrm{R}\text{, }&С=\frac{3x^{2}}{2}\text{ and }d=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Combine 2x^{3} and -2x^{3} to get 0.

$$dy=\int 3x\mathrm{d}x$$

The equation is in standard form.

$$dy=\frac{3x^{2}}{2}+С$$

Divide both sides by d.

$$\frac{dy}{d}=\frac{\frac{3x^{2}}{2}+С}{d}$$

Dividing by d undoes the multiplication by d.

$$y=\frac{\frac{3x^{2}}{2}+С}{d}$$