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-   -   stumped again! integral fraction (http://mymathforum.com/calculus/22771-stumped-again-integral-fraction.html)

 Ian McPherson November 22nd, 2011 01:17 PM

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?

 MarkFL November 22nd, 2011 01:20 PM

Re: stumped again! integral fraction

Looks good to me! :mrgreen:

 Ian McPherson November 22nd, 2011 01:22 PM

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!

 MarkFL November 22nd, 2011 03:47 PM

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:

$\frac{1}{8}\ln\|4x^2+1\|+C$ ?

 Ian McPherson November 22nd, 2011 08:17 PM

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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