Here's the question you clicked on:
kimmy0394
Evaluate the integral using the substitution rule. (Use C for the constant of integration.) Integration of 2y/((2+3y^2)^3) dy
\[\int\limits \frac{2y}{(2+3y^2)^3} dy\] \[ u=2+3y^2\] \[\frac{du}{dy}=6y \Rightarrow dy=\frac{du}{6y}\] \[\Rightarrow \int\limits \frac{2y}{(u)^3} \frac{du}{6y}\]You can do the rest
arent' the variables supposed to be the same?
i'm not getting the answer. i thought it would be 6y/(3(2+3y^2)^3) +C
It cancels to\[\frac{1}{3}\int\limits\frac{du}{u^3}=\frac{1}{3}\frac{1}{-4u^4}\]
oh i forgot to integrate. i just got so confused on the substitution. oops! :) thanks a bunch!
yeah, it was wrong. remember that 1/u^3 is the same as u^-3. thus the integration of u^-3 is u^-2 / -2 or 1/ (-2u^2)
Sorry about that- I half misread it as \[u^3\], not \[\frac{1}{u^3}\]. You're right