What is the integral from infinity to r of 1/(r^2) ?

The integral from infinity to r of 1/(r^2) is -1/r

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integral from infinity to r of 1/(r^2)

\int_{\infty}^{r} \frac{1}{r ^{2}} d r

- \frac{1}{r}

Solution steps

\int_{\infty}^{r} \frac{1}{r ^{2}} d r

Compute the indefinite integral:

\int \frac{1}{r ^{2}} d r = - \frac{1}{r} + C

Compute the boundaries:

\int_{\infty}^{r} \frac{1}{r ^{2}} d r = - \frac{1}{r} - 0

= - \frac{1}{r} - 0

Simplify

= - \frac{1}{r}

\int_{- 4}^{2} \frac{1}{x + 4} d x

\int_{0}^{4} - \frac{x}{2} d x

\int_{- 2}^{- 1} - x - 1 d x

\int_{1}^{2} (y - \frac{1}{y ^{2}}) d y

\int_{1}^{x^{2}} 1 + t d t

What is the integral from infinity to r of 1/(r^2) ?
The integral from infinity to r of 1/(r^2) is -1/r