integral of 1/(t^2sqrt(400-t^2))

Let's integrate:

\int(\frac{1}{t^{2} \cdot \sqrt{400 - t^{2}}})dt

We look for an antiderivative of the form

\frac{\sqrt{400 - t^{2}}}{t}

. The quotient rule and the derivative of the square root give

\frac{d}{d t} (\frac{\sqrt{400 - t^{2}}}{t}) = - \frac{400}{\sqrt{400 - t^{2}} \cdot t^{2}}

The integrand is

- \frac{1}{400}

times this derivative, so we multiply the antiderivative by

- \frac{1}{400}

This calculation is valid on each real interval where

400 - t^{2} > 0

The denominator also requires

t

≠ 0

- \frac{\sqrt{400 - t^{2}}}{400 t}

Indefinite integrals are defined up to an additive constant C, so this yields:

- \frac{\sqrt{400 - t^{2}}}{400 t} + C

▶Antiderivative:

- \frac{\sqrt{- t^{2} + 400}}{400 t} + C

, with C being any constant