Integrating a Rational Function: x/(x^2 + 5x + 6)

revlvr357
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hey guys, the function is...

(x dx) / (x^2 + 5x + 6)

please integrate it. please.

thanks so much,

Matt
 
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As it says in the guidelines when you joined PF just now, we don't do your homework for you. If you post some work of your own and show us where you are having problems, we should be able to offer some hints to keep you going.

So, can you figure out a function f(x) that you could differentiate to get something like the function you want to ingegrate? What techniques have you learned about to help you put fractions into forms that are easier to integrate?
 
ok, well so far I've learned 'u' substitution and integration by parts.

if i use 'u' substitution, u most likely = x^2 + 5x + 6. the demoninator could be simplified to (x+2)(x+3), but i don't think that would help much. if you use the 'u' i first suggested, du/dx = (2x + 5).

i can't seem to see how to get the 2x+ 5 to work as my (x dx)

as for integration by parts, i hardly understand that as we went over that lightly and will come back to it later.

so, this is where I'm stuck.

please help.
 
Maybe check out the techniques being discussed about halfway down this tutorial page:

http://www.karlscalculus.org/calc11_4.html
 
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ok, not going to lie, i read that entire page, and I'm still rather lost. seriously. i feel so stupid right now. i know I've done stuff like this before, but this is kill me.

can someone please give me a hint? just a hint?
 
Here's a hint -- cheat and plug the integral into Mathematica, and then look at the answer and differentiate it to see how it can become the integrand. Then think about what you would have to do with the integrand to get it into the original form of the question...

Mathematica said:
In[1]= Integrate[x/(x^2 + 5x + 6), x]

Out[1]= -2 Log[2 + x] + 3 Log[3 + x]
 
berkeman said:
Here's a hint -- cheat and plug the integral into Mathematica, and then look at the answer and differentiate it to see how it can become the integrand. Then think about what you would have to do with the integrand to get it into the original form of the question...

EDIT -- Oops, sorry for the potential confusion. In Mathematica, the "Log" function defaults to base e, hence it really is the ln() function. I never liked that about Mathematica. Obviously d ln(x) / dx = 1/x.
 
ok, well...i used your fist hint and i figured out it would be the natural log...
but anyway..

because of your hint, i was able to break the function into partial fractions and solve it for the answer mathematica spit out.

thank you very much.

matt
 
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