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MarcLeclair

  • 2 years ago

how do you do the following integral : int (4t / (1+6t^2)) I know the answer is 1/8 ln ( 1 +6t^2) but in the process, the answer shows the numerator becoming 32t and I have no idea where the 1/8 came from.

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  1. Outkast3r09
    • 2 years ago
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    \[\int \frac{4t}{1+6t^2}\]

  2. Outkast3r09
    • 2 years ago
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    \[4\int \frac{t}{1+6t^2}dt \] \[u=1+6t^2\] \[du=12tdt\] \[\frac{du}{12}=tdt\]

  3. MarcLeclair
    • 2 years ago
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    Ah you used substitution. Thanks a lot, makes muh more sense

  4. Outkast3r09
    • 2 years ago
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    ys and it's 1/3

  5. Outkast3r09
    • 2 years ago
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    wolframalpha is a good site to use to check

  6. campbell_st
    • 2 years ago
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    ok... so it looks like a log function to start s = 1 + 6u^2 ds/dt = 12t but you only need 4t so it becomes 1/3 of ds/dt so \[\int\limits \frac{\frac{1}{3} \times 12t}{1 + 6t^2} dt = \frac{1}{3} \int\limits \frac{4t}{1 + 6t^2} dt\]

  7. Outkast3r09
    • 2 years ago
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    the 1/8 might be a typo

  8. MarcLeclair
    • 2 years ago
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    Thanks to both of you :) help is greatly appreciated haha. And it looks like it. 1/3 does seem right!

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