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anonymous

  • 5 years ago

Find the integral:

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  1. anonymous
    • 5 years ago
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    t^2 + 4 / t^3 +12t +8 dt

  2. anonymous
    • 5 years ago
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    Do you know how an integral works?

  3. anonymous
    • 5 years ago
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    They confuse me.

  4. anonymous
    • 5 years ago
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    Is it \[\int\limits_{}^{}{t^2+4 \over t^3+12t+8} dt?\]

  5. anonymous
    • 5 years ago
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    @AnwarAL it is what you posted. I think you can do u-substition for the integral.

  6. anonymous
    • 5 years ago
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    OK let's call the integral I .. if you look closely you will see that The numerator is the derivative of the denominator after multiplying it (The numerator) by 3.. SO, multiply The numerator by 3 and divide outside the integral by 3, you get: \[I={1 \over 3}\int\limits_{}^{}{3t^2+12 \over t^3+12t+8}dt\] now you can see that The numerator is exactly the derivative of the numerator, then: \[I={1 \over 3} \ln \left| t^3+12t+8 \right|+c\]

  7. anonymous
    • 5 years ago
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    you can do substitution u=t^3+12t+8 as smorlaz said.. the way I used looks faster and easier to me. They both lead to the same result.

  8. anonymous
    • 5 years ago
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    does that make sense to you?

  9. anonymous
    • 5 years ago
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    Wait a minute, why can't you just seperate the terms and integrate individually?

  10. anonymous
    • 5 years ago
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    Or did I read the problem incorrectly?

  11. anonymous
    • 5 years ago
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    ^^ I think you did :)

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