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anonymous
 5 years ago
∫[(e^x)/(e^(2x)+3e^x+2),]
anonymous
 5 years ago
∫[(e^x)/(e^(2x)+3e^x+2),]

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lokisan u are you replying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The 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,\[I=\log \frac{e^x+1}{e^x+2}+c\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you. this looks so similar to what we are doing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0np  I'd appreciate another fan :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Whenever you see a polynomial in the denominator, try to factor it. Then try to see if you can use partial fraction decomposition.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dichalao, what the hell's up with your professor giving you that question? It's an algebraic pain.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0How many more questions do you have?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You may want to spread those questions around (i.e. keep making new posts) so others can help too.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how can i type the equation easily like you did in the reply?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0There's a button below, "Equation". You can enter things in there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry but can you show the decomposition part?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Loki do u have time to help me out on that problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no in the first answer. you said a=1 b=1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes dichalao...just let me finish here and I'll go back.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0In 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.
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