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anonymous

  • 4 years ago

Calculus II - Trig Integration Hey all, just wanted to confirm one answer and likely ask another question in a different thread; \[\int\limits_{}^{} 1/(x^2+6x+12) dx \] \[(x^2+6x+9) = (x+3)^2 + 3 \] \[\int\limits_{}^{}1/a^2 + u^2 = (1/a)*\tan^-1(u/a) + C \] \[1/\sqrt(3) * \tan^-1 ((x+3)/\sqrt(3)) + C \] Can anyone confirm that answer?

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  1. anonymous
    • 4 years ago
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    ya

  2. anonymous
    • 4 years ago
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    Great, thanks! Now, I have another trig integration which I am actually just stuck on. Can I ask that here or should I post a new thread?

  3. anonymous
    • 4 years ago
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    Ill start typing it out here and probably post here and a new thread.

  4. anonymous
    • 4 years ago
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    Evaluate the integral; \[\int\limits(1/(x*\sqrt(x^8-9) dx\] So I know this is in the form \[\int\limits du/u*\sqrt(u^2 - a^2) = (1/a) \sec^-1(|u/a|) + C\] But I am not sure how to get that X and X^8 out of there. I know if the X was X^4 then X^8 would be (X^4)^2 but I am not really sure what to do. Any help starting this problem out would be great!

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