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December 10th, 2017, 09:47 AM   #1
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Antiderivative

Hi guys,
Any idea about how I can take antiderivative of (u^2-3)/(u^2+3)?
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December 10th, 2017, 10:19 AM   #2
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$\dfrac{u^2+3-6}{u^2+3} = 1 - \dfrac{6}{u^2+3}$

$\displaystyle \int 1 - \dfrac{6}{u^2+3} \, du$

take it from here?

... antiderivative of the 2nd term will involve an arctangent
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December 10th, 2017, 04:36 PM   #3
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Quote:
Originally Posted by skeeter View Post
$\dfrac{u^2+3-6}{u^2+3} = 1 - \dfrac{6}{u^2+3}$

$\displaystyle \int 1 - \dfrac{6}{u^2+3} \, du$

take it from here?

... antiderivative of the 2nd term will involve an arctangent
Preview your latex - it is not coming out right.
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December 10th, 2017, 04:38 PM   #4
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Quote:
Originally Posted by mathman View Post
Preview your latex - it is not coming out right.
looks fine to me ... what do you see?
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January 14th, 2018, 09:06 PM   #5
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Quote:
Originally Posted by skeeter View Post
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.
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January 15th, 2018, 12:47 PM   #6
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Quote:
Originally Posted by mathman View Post
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.
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January 15th, 2018, 12:51 PM   #7
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$\frac{u^2+3-6}{u^2+3} = 1 - \frac{6}{u^2+3}$

$\displaystyle \int 1 - \frac{6}{u^2+3} \, du$

take it from here?

... antiderivative of the 2nd term will involve an arctangent

(I changed dfrac to frac - now it is readable.)
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January 15th, 2018, 03:09 PM   #8
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Perhaps you should refresh with shift-f5 to reload all the cached material (or delete the cache and reload the page).
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January 15th, 2018, 06:51 PM   #9
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Quote:
Originally Posted by mathman View Post
$\frac{u^2+3-6}{u^2+3} = 1 - \frac{6}{u^2+3}$

$\displaystyle \int 1 - \frac{6}{u^2+3} \, du$

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
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January 16th, 2018, 08:26 AM   #10
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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!)
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Last edited by greg1313; January 17th, 2018 at 12:53 AM.
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