Solve ∫ wrt x/sqrt[3]{x}=3x^2/3/2+c

$$\int \frac { d x } { \sqrt[ 3 ] { x } } = \frac { 3 x ^ { 2 / 3 } } { 2 } + c$$

Solve for c

$$c=С$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$2\int \frac{1}{\sqrt[3]{x}}\mathrm{d}x=3x^{\frac{2}{3}}+2c$$

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

$$3x^{\frac{2}{3}}+2c=2\int \frac{1}{\sqrt[3]{x}}\mathrm{d}x$$

Subtract 3x^{\frac{2}{3}} from both sides.

$$2c=2\int \frac{1}{\sqrt[3]{x}}\mathrm{d}x-3x^{\frac{2}{3}}$$

The equation is in standard form.

$$2c=2С+\frac{2x}{\sqrt[3]{x}}-3x^{\frac{2}{3}}$$

Divide both sides by 2.

$$\frac{2c}{2}=\frac{2С-x^{\frac{2}{3}}}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$c=\frac{2С-x^{\frac{2}{3}}}{2}$$

Divide 2С-x^{\frac{2}{3}} by 2.

$$c=-\frac{x^{\frac{2}{3}}}{2}+С$$

Solve for x

$$x\neq 0$$

$$c=С\text{ and }x\neq 0$$