Solve ∫ sqrt{ax wrt x=frac{2xsqrt{ax}}{3}+c

$$\int \sqrt { a x } d x = \frac { 2 x \sqrt { a x } } { 3 } + c$$

Solve for c

$$c=С$$

$$\left(x\leq 0\text{ and }a<0\right)\text{ or }\left(x\geq 0\text{ and }a>0\right)$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 3.

$$3\int \sqrt{ax}\mathrm{d}x=2x\sqrt{ax}+3c$$

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

$$2x\sqrt{ax}+3c=3\int \sqrt{ax}\mathrm{d}x$$

Subtract 2x\sqrt{ax} from both sides.

$$3c=3\int \sqrt{ax}\mathrm{d}x-2x\sqrt{ax}$$

The equation is in standard form.

$$3c=3x\sqrt{ax}-2x\sqrt{ax}+3С$$

Divide both sides by 3.

$$\frac{3c}{3}=\frac{x\sqrt{ax}+3С}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$c=\frac{x\sqrt{ax}+3С}{3}$$

Divide \sqrt{ax}x+3С by 3.

$$c=\frac{x\sqrt{ax}}{3}+С$$

Solve for a

$$\left\{\begin{matrix}a=\frac{С_{1}^{2}+9c^{2}-Сc}{x^{3}}\text{, }&\left(c\leq С\text{ and }x<0\right)\text{ or }\left(c\geq С_{1}\text{ and }x>0\right)\\a\in \mathrm{R}\text{, }&x=0\text{ and }c=С\end{matrix}\right.$$