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could someone please check my answer for integrating v^2 dv/dt = (2+t)^3? watch

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    I have  v^2 \frac {dv}{dt} = (2+t)^3

    \int v^2 dv = \int (2+t)^3 dt

     \frac {v^3}{3} = \frac {(2+t)^4}{4}*\frac{1}{3}+A
    \frac {v^3}{3} = \frac {1}{12}*(2+t)^4+A

    Can't see how I would get the answer my book has:
    4v^3 = 3(2+t)^4+A
    even if I multiply everything by 12 which would give 4v^3 = (2+t)^4+a.

    I used the rule  \int (ax +b)^n = \frac {(ax+b)^{n+1}}{n+1}*\frac{1}{a}+  c not sure if that was correct because there is constant where ax is?

    Thanks
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    Why did you multiply by \dfrac{1}{3} in your right-hand integral? When using the rule

    \displaystyle \int (ax+b)^n\, dx = \dfrac{(ax+b)^{n+1}}{n+1} \times \dfrac{1}{a} + c

    a is just the coefficient of x, which in this case is 1 (not 3).
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    (Original post by nuodai)
    Why did you multiply by \dfrac{1}{3} in your right-hand integral? When using the rule

    \displaystyle \int (ax+b)^n\, dx = \dfrac{(ax+b)^{n+1}}{n+1} \times \dfrac{1}{a} + c

    a is just the coefficient of x, which in this case is 1 (not 3).
    course it is, thanks! I was using n for some reason. Thank you!
 
 
 
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