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anonymous

  • 4 years ago

evaluate the following: integral S (e^3x)/((e^6x)+1) dx (hint: u=e^3x)

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  1. Mr.Math
    • 4 years ago
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    \[u=e^{3x} \implies du=3e^{3x}dx\] \[I=\frac{1}{3}\int \frac{du}{u^2+1}=\frac{1}{3}\tan^{-1}u+c=\frac{1}{3}\tan^{-1}(e^{3x})+c.\]

  2. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}\frac{e^{3x}}{e^{6x}+1} dx\] Thinking.... \[e^{3x}=\tan(\theta)\] => dang it

  3. myininaya
    • 4 years ago
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    great job math lol

  4. Mr.Math
    • 4 years ago
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    Obviously \(I\) is the integral you gave.

  5. Mr.Math
    • 4 years ago
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    @myin: My substitution was a hint from the question :-D

  6. myininaya
    • 4 years ago
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    \[\int\limits_{}^{}\frac{ \frac{1}{3} \sec^2(\theta)}{\tan^2(\theta)+1} d \theta\]

  7. myininaya
    • 4 years ago
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    i like my sub better

  8. myininaya
    • 4 years ago
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    it contains both subs

  9. anonymous
    • 4 years ago
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    haha that's awesome. i knew it was (1/3)arctan(e3x) + C. i was just seeing if this actually worked :)

  10. anonymous
    • 4 years ago
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    \[\int\limits \frac{e^{3 x}}{1+e^{6 x}} \, dx=\frac{1}{3} \text{ArcTan}\left[e^{3 x}\right]+c \]

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