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

1. anonymous

lokisan u are you replying?

2. anonymous

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

is this the answer?

4. anonymous

Yes

5. anonymous

thank you. this looks so similar to what we are doing

6. anonymous

np - I'd appreciate another fan :P

7. anonymous

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

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

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

dichalao, what the hell's up with your professor giving you that question? It's an algebraic pain.

11. anonymous

she's evil

12. anonymous

How many more questions do you have?

13. anonymous

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

You may want to spread those questions around (i.e. keep making new posts) so others can help too.

15. anonymous

do you know how can i type the equation easily like you did in the reply?

16. anonymous

There's a button below, "Equation". You can enter things in there.

17. anonymous

sorry but can you show the decomposition part?

18. anonymous

Loki do u have time to help me out on that problem?

19. anonymous

no in the first answer. you said a=1 b=-1

20. anonymous

Yes dichalao...just let me finish here and I'll go back.

21. anonymous

22. anonymous

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

thanks