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November 22nd, 2011, 01:17 PM   #1
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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?
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November 22nd, 2011, 01:20 PM   #2
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Re: stumped again! integral fraction

Looks good to me!
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November 22nd, 2011, 01:22 PM   #3
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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!
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November 22nd, 2011, 03:47 PM   #4
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Re: stumped again! integral fraction

You want the one that interprets as:



Do you understand why we do not need to write:

?
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November 22nd, 2011, 08:17 PM   #5
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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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