anonymous
  • anonymous
\[\int\limits_{}^{} (6x-11 / x+2) dx\]
Mathematics
jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
nice job with the equation writing skill...kudos ;)
amistre64
  • amistre64
frac{top}{bottom} makes a clean fraction
anonymous
  • anonymous
You need to substitute x for x=u+2 and then solve (dx=du since d/dy of +2 is zero).

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anonymous
  • anonymous
Err... that's not d/dy of +2 but the derivative of +2
amistre64
  • amistre64
\[\frac{top}{australia}\]
amistre64
  • amistre64
\[\int \frac{6x-11}{x+2}dx\]
anonymous
  • anonymous
thanks for the lesson in writing equations >:(
anonymous
  • anonymous
Do you get what I wrote msunprecedented?
anonymous
  • anonymous
no, i did not
amistre64
  • amistre64
lol.... I was impressed by your talent ...still am :)
amistre64
  • amistre64
do you want the integral of 3 seperate terms; or is what I posted a correct interpretation of it?
anonymous
  • anonymous
no its correct \[\int\limits_{}^{} \frac {6x-11} {x+2} dx\]
amistre64
  • amistre64
... factoring might be possible but we would have to find out what value had been canceled before hand.... and then do partial decomposition as a possibility
anonymous
  • anonymous
Actually, this is really simple if you view the fraction like this: \[6x/(x+2) - (11/x+2)\] Now that it's in this form, you split the integral into two : \[\int\limits_{?}^{?}6x/(x+2)dx - \int\limits_{?}^{?}11/(6x+2)dx\] This should be easier to integrate... let me know if you need more help
amistre64
  • amistre64
the left is easy now :) the right still has a straggler.... that "x" up top can get bothersome
amistre64
  • amistre64
and by left I mean right ;)
amistre64
  • amistre64
would the uv substitution work well there?
anonymous
  • anonymous
uv substitution wouldn't work because I don't think you could factor x+2. (I can't see a way to factor anyway)
amistre64
  • amistre64
6x goes to zero pretty quick like; but I get ln(x+2) and get lost on integrateing that
amistre64
  • amistre64
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anonymous
  • anonymous
Hmm.. I think integration by substitution works here. If you let u=x+2, you can get the right equal to:\[\int\limits_{?}^{?} (6u-12)/u *du\] Then you split the fraction one more time
anonymous
  • anonymous
I mean the left when I say the "right" above
amistre64
  • amistre64
lol..... stage left ...STAGE left!! :)
anonymous
  • anonymous
Yikes! I'm soo confused! U-sub here? Integration by parts there? :(
anonymous
  • anonymous
msunprecedented, have you learnt how to do integration by substitution yet?
anonymous
  • anonymous
I believe not.

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