# Math Help - help with a tough integration

1. ## help with a tough integration

i need help solving this integral to use it in a physics problem.

$\int(dx/(10.2-x)^2)$

$\int(dx/(10.2-x)^2)$
Let $y = 10.2 - x$ ==> $dy = -dx$
$\int \frac{dx}{(10.2-x)^2} = \int \frac{-dy}{y^2} = \frac{1}{y} + C = \frac{1}{10.2 - x} + C$