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kimmy0394

  • 3 years ago

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

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  1. henpen
    • 3 years ago
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    \[\int\limits \frac{2y}{(2+3y^2)^3} dy\] \[ u=2+3y^2\] \[\frac{du}{dy}=6y \Rightarrow dy=\frac{du}{6y}\] \[\Rightarrow \int\limits \frac{2y}{(u)^3} \frac{du}{6y}\]You can do the rest

  2. kimmy0394
    • 3 years ago
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    arent' the variables supposed to be the same?

  3. kimmy0394
    • 3 years ago
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    i'm not getting the answer. i thought it would be 6y/(3(2+3y^2)^3) +C

  4. henpen
    • 3 years ago
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    It cancels to\[\frac{1}{3}\int\limits\frac{du}{u^3}=\frac{1}{3}\frac{1}{-4u^4}\]

  5. kimmy0394
    • 3 years ago
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    oh i forgot to integrate. i just got so confused on the substitution. oops! :) thanks a bunch!

  6. kimmy0394
    • 3 years ago
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    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
    • 3 years ago
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    Sorry about that- I half misread it as \[u^3\], not \[\frac{1}{u^3}\]. You're right

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