i need help solving this integral to use it in a physics problem. $\displaystyle \int(dx/(10.2-x)^2)$
Follow Math Help Forum on Facebook and Google+
Originally Posted by Chadly724 i need help solving this integral to use it in a physics problem. $\displaystyle \int(dx/(10.2-x)^2)$ Let $\displaystyle y = 10.2 - x$ ==> $\displaystyle dy = -dx$ So $\displaystyle \int \frac{dx}{(10.2-x)^2} = \int \frac{-dy}{y^2} = \frac{1}{y} + C = \frac{1}{10.2 - x} + C$ -Dan
View Tag Cloud