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 December 10th, 2017, 09:47 AM #1 Member   Joined: Apr 2017 From: New York Posts: 63 Thanks: 6 Antiderivative Hi guys, Any idea about how I can take antiderivative of (u^2-3)/(u^2+3)?
 December 10th, 2017, 10:19 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,725 Thanks: 1376 $\dfrac{u^2+3-6}{u^2+3} = 1 - \dfrac{6}{u^2+3}$ $\displaystyle \int 1 - \dfrac{6}{u^2+3} \, du$ take it from here? ... antiderivative of the 2nd term will involve an arctangent
December 10th, 2017, 04:36 PM   #3
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Quote:
 Originally Posted by skeeter $\dfrac{u^2+3-6}{u^2+3} = 1 - \dfrac{6}{u^2+3}$ $\displaystyle \int 1 - \dfrac{6}{u^2+3} \, du$ take it from here? ... antiderivative of the 2nd term will involve an arctangent
Preview your latex - it is not coming out right.

December 10th, 2017, 04:38 PM   #4
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Quote:
 Originally Posted by mathman Preview your latex - it is not coming out right.
looks fine to me ... what do you see?

January 14th, 2018, 09:06 PM   #5
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Quote:
 Originally Posted by skeeter looks fine to me ... what do you see?
He is referring to where you replace -3 with 3-6. It took me a second to understand what you did there as well.

Last edited by skipjack; January 15th, 2018 at 12:01 AM.

January 15th, 2018, 12:47 PM   #6
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Quote:
 Originally Posted by mathman Preview your latex - it is not coming out right.
Where the formula is supposed to be I get "Undefined control sequence ...", indicating that something is not working to translate the latex.

 January 15th, 2018, 12:51 PM #7 Global Moderator   Joined: May 2007 Posts: 6,454 Thanks: 567 $\frac{u^2+3-6}{u^2+3} = 1 - \frac{6}{u^2+3}$ $\displaystyle \int 1 - \frac{6}{u^2+3} \, du$ take it from here? ... antiderivative of the 2nd term will involve an arctangent (I changed dfrac to frac - now it is readable.)
 January 15th, 2018, 03:09 PM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,234 Thanks: 2412 Math Focus: Mainly analysis and algebra Perhaps you should refresh with shift-f5 to reload all the cached material (or delete the cache and reload the page).
January 15th, 2018, 06:51 PM   #9
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Quote:
 Originally Posted by mathman $\frac{u^2+3-6}{u^2+3} = 1 - \frac{6}{u^2+3}$ $\displaystyle \int 1 - \frac{6}{u^2+3} \, du$ take it from here? ... antiderivative of the 2nd term will involve an arctangent (I changed dfrac to frac - now it is readable.)
Pet peeve warning! Pet peeve! Danger, Will Robinson! Danger!

The integral really should be written as
$\displaystyle \int \left ( 1 - \frac{6}{u^2+3} \right ) ~ du$

I'll grant that there is only one way to take the integral as it was written since $\displaystyle \int 1$ means nothing. But I still feel it needs to be there.

Pet peeve over with.

-Dan

 January 16th, 2018, 08:26 AM #10 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,967 Thanks: 807 I disagree. The function $\displaystyle 1-\frac{6}{u^2+ 3}$ lies between $\displaystyle \int$ and $\displaystyle dx$. The parentheses are unnecessary. The $\displaystyle \int$ and $\displaystyle dx$ take their place. (My pet peeve can beat your pet peeve!) Thanks from topsquark and v8archie Last edited by greg1313; January 17th, 2018 at 12:53 AM.

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