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anonymous

  • 5 years ago

∫[(e^x)/(e^(2x)+3e^x+2),]

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  1. anonymous
    • 5 years ago
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    lokisan u are you replying?

  2. anonymous
    • 5 years ago
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    The denominator can be factored as\[(e^x+1)(e^x+2)\]If you then set \[u=e^x \rightarrow du=e^x dx\]you'll end up with\[I=\int\limits_{}{}\frac{e^x}{e^{2x}+3e^x+2}dx=\int\limits_{}{}\frac{du}{(u+1)(u+2)}\]You can use partial fraction decomposition on this last expression:\[\frac{1}{(u+1)(u+2)}=\frac{A}{u+1}+\frac{B}{u+2}\]to find\[A=1, B=-1\]Your integral is then\[I=\int\limits_{}{}\frac{1}{u+1}-\frac{1}{u+2}du = \log (u+1)-\log(u+2)+c\]Substituting back,\[I=\log \frac{e^x+1}{e^x+2}+c\]

  3. anonymous
    • 5 years ago
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    is this the answer?

  4. anonymous
    • 5 years ago
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    Yes

  5. anonymous
    • 5 years ago
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    thank you. this looks so similar to what we are doing

  6. anonymous
    • 5 years ago
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    np - I'd appreciate another fan :P

  7. anonymous
    • 5 years ago
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    Whenever you see a polynomial in the denominator, try to factor it. Then try to see if you can use partial fraction decomposition.

  8. anonymous
    • 5 years ago
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    how do you become a fan? you are a lifesaver. My evil calculas teacher gave us a take home test even i missed the whole week. thank you and can you help with other problems?

  9. anonymous
    • 5 years ago
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    there should be a link next to my name - a blue line saying, "Become a fan". If it's not there, maybe try refreshing you page.

  10. anonymous
    • 5 years ago
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    dichalao, what the hell's up with your professor giving you that question? It's an algebraic pain.

  11. anonymous
    • 5 years ago
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    she's evil

  12. anonymous
    • 5 years ago
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    How many more questions do you have?

  13. anonymous
    • 5 years ago
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    I have 10 out of which i have done like 3 . 3 of them are set up only using fraction decomposition. and 2 are solving using decomposition

  14. anonymous
    • 5 years ago
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    You may want to spread those questions around (i.e. keep making new posts) so others can help too.

  15. anonymous
    • 5 years ago
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    do you know how can i type the equation easily like you did in the reply?

  16. anonymous
    • 5 years ago
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    There's a button below, "Equation". You can enter things in there.

  17. anonymous
    • 5 years ago
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    sorry but can you show the decomposition part?

  18. anonymous
    • 5 years ago
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    Loki do u have time to help me out on that problem?

  19. anonymous
    • 5 years ago
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    no in the first answer. you said a=1 b=-1

  20. anonymous
    • 5 years ago
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    Yes dichalao...just let me finish here and I'll go back.

  21. anonymous
    • 5 years ago
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  22. anonymous
    • 5 years ago
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    In partial fraction decomposition, you set it out like that when you have linear factors in the bottom. You then multiply both sides by (u+1)(u+2) and you'll end up with 1 on the left and what you see on the right. You need to find A and B that will make the statement true; that is, by comparing coefficients on the left and right.

  23. anonymous
    • 5 years ago
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    thanks

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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