My Math Forum stumped again! integral fraction

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 November 22nd, 2011, 01:17 PM #1 Newbie   Joined: Oct 2011 Posts: 26 Thanks: 0 stumped again! integral fraction integral (x / (4x^2 +1) ) dx u is 4x^2 + 1, du is 8x therefore, multiply numerator x by 8, and put 1/8 outside of integral to get 1/8 ln |4x^2 + 1| + c This is just a weird problem. I haven't seen one like it, and I tried my best. Can you please explain any errors I've made?
 November 22nd, 2011, 01:20 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: stumped again! integral fraction Looks good to me!
 November 22nd, 2011, 01:22 PM #3 Newbie   Joined: Oct 2011 Posts: 26 Thanks: 0 Re: stumped again! integral fraction The professor sent us home with a practice test in anticipation of our final normal test, because it's kinda a big deal :P and he did it in multiple choice form to make it harder I guess. So none of them look right, but one is 1/8ln(4x^2+1) + c. Would that be the same answer? Because none of these is also an answer, I'd hate to not know if it was and miss that on the test!
 November 22nd, 2011, 03:47 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: stumped again! integral fraction You want the one that interprets as: $\frac{1}{8}\ln$$4x^2+1$$+C$ Do you understand why we do not need to write: $\frac{1}{8}\ln\|4x^2+1\|+C$ ?
 November 22nd, 2011, 08:17 PM #5 Newbie   Joined: Oct 2011 Posts: 26 Thanks: 0 Re: stumped again! integral fraction Oh yes I do now! The absolute value bars are there to make it positive, but the 4x^2 ensures that any number being squared will turn out positive, thus meaning that the absolute value bars are not necessary!

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