Evaluate integral of 1/(r^2) with respect to r

$\int \frac{1}{r^{2}} d r$

Apply basic rules of exponents.

Tap for more steps…

Move $r^{2}$ out of the denominator by raising it to the $- 1$ power.

$\int {(r^{2})}^{- 1} d r$

Multiply the exponents in ${(r^{2})}^{- 1}$ .

Tap for more steps…

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\int r^{2 \cdot - 1} d r$

Multiply $2$ by $- 1$ .

$\int r^{- 2} d r$

By the Power Rule, the integral of $r^{- 2}$ with respect to $r$ is $- r^{- 1}$ .

$- r^{- 1} + C$

Rewrite $- r^{- 1} + C$ as $- \frac{1}{r} + C$ .

$- \frac{1}{r} + C$

Evaluate integral of 1/(r^2) with respect to r

Do you need help with math?