Solve ∫ x wrt x=x^2/x+c

$$\int x d x = \frac { x ^ { 2 } } { x } + c$$

Solve for c

$$c=\frac{x^{2}}{2}-x+С$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$x\int x\mathrm{d}x=x^{2}+xc$$

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

$$x^{2}+xc=x\int x\mathrm{d}x$$

Subtract x^{2} from both sides.

$$xc=x\int x\mathrm{d}x-x^{2}$$

The equation is in standard form.

$$xc=x\left(\frac{x^{2}}{2}+С\right)-x^{2}$$

Divide both sides by x.

$$\frac{xc}{x}=\frac{x\left(\frac{x^{2}}{2}-x+С\right)}{x}$$

Dividing by x undoes the multiplication by x.

$$c=\frac{x\left(\frac{x^{2}}{2}-x+С\right)}{x}$$

Divide x\left(\frac{x^{2}}{2}+С-x\right) by x.

$$c=\frac{x^{2}}{2}-x+С$$

Solve for x

$$\left\{\begin{matrix}x=-\sqrt{С+2c}+1\text{, }&\left(С<0\text{ or }c\neq \frac{1-С_{1}}{2}\right)\text{ and }c\geq С_{2}\\x=\sqrt{С+2c}+1\text{, }&c\geq С\end{matrix}\right.$$