Solve ∫ sqrt[3]{3t}dt=(3t)^4/2/4tc

$$\int \sqrt[ 3 ] { 3 t } d t = \frac { ( 3 t ) ^ { 4 / 2 } } { 4 } t c$$

Solve for c

$$\left\{\begin{matrix}c=\frac{3^{\frac{4}{3}}}{9t^{\frac{5}{3}}}+\frac{4С}{9t^{3}}\text{, }&t\neq 0\\c\in \mathrm{R}\text{, }&С=0\text{ and }t=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4.

$$4\int \sqrt[3]{3t}\mathrm{d}t=\left(3t\right)^{\frac{4}{2}}tc$$

Divide 4 by 2 to get 2.

$$4\int \sqrt[3]{3t}\mathrm{d}t=\left(3t\right)^{2}tc$$

Expand \left(3t\right)^{2}.

$$4\int \sqrt[3]{3t}\mathrm{d}t=3^{2}t^{2}tc$$

Calculate 3 to the power of 2 and get 9.

$$4\int \sqrt[3]{3t}\mathrm{d}t=9t^{2}tc$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$4\int \sqrt[3]{3t}\mathrm{d}t=9t^{3}c$$

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

$$9t^{3}c=4\int \sqrt[3]{3t}\mathrm{d}t$$

The equation is in standard form.

$$9t^{3}c=4\sqrt[3]{3}t^{\frac{4}{3}}+4С$$

Divide both sides by 9t^{3}.

$$\frac{9t^{3}c}{9t^{3}}=\frac{\frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С}{9t^{3}}$$

Dividing by 9t^{3} undoes the multiplication by 9t^{3}.

$$c=\frac{\frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С}{9t^{3}}$$

Divide \frac{4\times \left(3t\right)^{\frac{4}{3}}}{3}+4С by 9t^{3}.

$$c=\frac{4\left(\frac{\left(3t\right)^{\frac{4}{3}}}{3}+С\right)}{9t^{3}}$$