## kimmy0394 2 years ago Evaluate the integral using the substitution rule. (Use C for the constant of integration.) Integration of 2y/((2+3y^2)^3) dy

1. henpen

$\int\limits \frac{2y}{(2+3y^2)^3} dy$ $u=2+3y^2$ $\frac{du}{dy}=6y \Rightarrow dy=\frac{du}{6y}$ $\Rightarrow \int\limits \frac{2y}{(u)^3} \frac{du}{6y}$You can do the rest

2. kimmy0394

arent' the variables supposed to be the same?

3. kimmy0394

i'm not getting the answer. i thought it would be 6y/(3(2+3y^2)^3) +C

4. henpen

It cancels to$\frac{1}{3}\int\limits\frac{du}{u^3}=\frac{1}{3}\frac{1}{-4u^4}$

5. kimmy0394

oh i forgot to integrate. i just got so confused on the substitution. oops! :) thanks a bunch!

6. kimmy0394

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)

7. henpen

Sorry about that- I half misread it as $u^3$, not $\frac{1}{u^3}$. You're right