Solve 2 ∫ wrt x/x^2=-1/x+c

$$2 \int \frac { d x } { x ^ { 2 } } = - \frac { 1 } { x } + c$$

Solve for c

$$c=-\frac{1}{x}+С$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by x.

$$2\int \frac{1}{x^{2}}\mathrm{d}xx=-1+xc$$

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

$$-1+xc=2\int \frac{1}{x^{2}}\mathrm{d}xx$$

Add 1 to both sides.

$$xc=2\int \frac{1}{x^{2}}\mathrm{d}xx+1$$

The equation is in standard form.

$$xc=Сx-1$$

Divide both sides by x.

$$\frac{xc}{x}=\frac{Сx-1}{x}$$

Dividing by x undoes the multiplication by x.

$$c=\frac{Сx-1}{x}$$

Divide -1+Сx by x.

$$c=-\frac{1}{x}+С$$

Solve for x

$$x=\frac{1}{С-c}$$

$$c\neq С$$