# help with a tough integration

• January 29th 2007, 06:34 PM
help with a tough integration
i need help solving this integral to use it in a physics problem.

$\int(dx/(10.2-x)^2)$
• January 29th 2007, 06:51 PM
topsquark
Quote:

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