Solve ∫ 2dt/t^2=-2/t+C

$$\int \frac { 2 d t } { t ^ { 2 } } = - \frac { 2 } { t } + C$$

Solve for C

$$C=С$$

$$t\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by t.

$$t\int \frac{2}{t^{2}}\mathrm{d}t=-2+tC$$

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

$$-2+tC=t\int \frac{2}{t^{2}}\mathrm{d}t$$

Add 2 to both sides.

$$tC=t\int \frac{2}{t^{2}}\mathrm{d}t+2$$

The equation is in standard form.

$$tC=Сt$$

Divide both sides by t.

$$\frac{tC}{t}=\frac{Сt}{t}$$

Dividing by t undoes the multiplication by t.

$$C=\frac{Сt}{t}$$

Divide Сt by t.

$$C=С$$

Solve for t

$$t\neq 0$$

$$C=С\text{ and }t\neq 0$$