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^23)/(u^2+3)? 
December 10th, 2017, 10:19 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 2,725 Thanks: 1376 
$\dfrac{u^2+36}{u^2+3} = 1  \dfrac{6}{u^2+3}$ $\displaystyle \int 1  \dfrac{6}{u^2+3} \, du$ take it from here? ... antiderivative of the 2nd term will involve an arctangent 
December 10th, 2017, 04:36 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,454 Thanks: 567  
December 10th, 2017, 04:38 PM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 2,725 Thanks: 1376  
January 14th, 2018, 09:06 PM  #5 
Newbie Joined: Jan 2018 From: somewhere Posts: 14 Thanks: 2 Math Focus: Algebraic Number Theory / Differential Fork Theory  He is referring to where you replace 3 with 36. 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 
Global Moderator Joined: May 2007 Posts: 6,454 Thanks: 567  
January 15th, 2018, 12:51 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,454 Thanks: 567 
$\frac{u^2+36}{u^2+3} = 1  \frac{6}{u^2+3}$ $\displaystyle \int 1  \frac{6}{u^2+3} \, du$ 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 shiftf5 to reload all the cached material (or delete the cache and reload the page).

January 15th, 2018, 06:51 PM  #9  
Math Team Joined: May 2013 From: The Astral plane Posts: 1,689 Thanks: 670 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
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!) Last edited by greg1313; January 17th, 2018 at 12:53 AM. 

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