anonymous
  • anonymous
I know that similar functions can have very different antiderivatives. For example, even though int((x)/((x^4)+1)) dx looks very similar to int(((x^4)+1)/(x)), their solutions are very different. How can I prove that I'm right by solving both?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
you wanna indefinite integral both of them right?
anonymous
  • anonymous
Yes.
amistre64
  • amistre64
the last one just split the fraction: x^4/x +1/x and integrate

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amistre64
  • amistre64
\[\int\limits_{} \frac{x^4}{x} + \frac{1}{x} dx\]
anonymous
  • anonymous
Right...
amistre64
  • amistre64
the second one we can try subsituting u=x^4+1
amistre64
  • amistre64
by second I mean first :)
anonymous
  • anonymous
I gotcha :) I figured as much...first and second...same thing. right? haha!
anonymous
  • anonymous
so, on the first one you end up with int(1/u)du?
anonymous
  • anonymous
\[\int\limits 1/u\]
amistre64
  • amistre64
\[\int\limits_{} \frac{1}{4u(u-1)^{1/2}}\]this is what I get :)
amistre64
  • amistre64
..du
amistre64
  • amistre64
put the u under the radial and take out the 1/4
amistre64
  • amistre64
sounds good, looks bad; decompose the fraction :)
amistre64
  • amistre64
\[\frac{1}{4}\int\limits_{}\frac{1}{u}du - \int\limits_{} \frac{1}{(u+1)^{1/2}}du\]
amistre64
  • amistre64
umm... what did I do lol...hold up on that
amistre64
  • amistre64
yeah, thats right, I couldnt recall where I got (u-1)^1/2 from
anonymous
  • anonymous
Yeah, I was about to say that you were correct. So, would you mind walking me through how to solve the second one?
amistre64
  • amistre64
and by second do you mean first?
anonymous
  • anonymous
Haha! yes, the first one we talked about anyway.
amistre64
  • amistre64
ok.... whenever we have 2 numbers above a single term; we can split that up to its baser seqments: x^4 + 1 x^4 1 ------- = ---- + ---- right? x x x
amistre64
  • amistre64
like denominators simply merge and you work the tops; well, they can be split up just as easily
anonymous
  • anonymous
yes.
amistre64
  • amistre64
we get: x^3 + 1/x which integrates to: x^4/4 + ln(x) + C
anonymous
  • anonymous
Oh! I understand! Thank you so much. Now, do we need to add a plus c to the second one we did?
amistre64
  • amistre64
whenever we have a inegral with no limits attached to it; we have to include the "constant of integration"...so yes. +C is just a place holder till we get more information about it to anchor it to the graph.
amistre64
  • amistre64
was the decomposition one right? or do we know?
anonymous
  • anonymous
I don't think we know, unless we can infer that it is from the original problem I gave.
amistre64
  • amistre64
let me redo that one on paper to see if I got it right....
amistre64
  • amistre64
running into problems with it ...
amistre64
  • amistre64
i might just be over thinking it, let me try trig substitution...
anonymous
  • anonymous
I didn't really notice anything wrong, but I could just be trying to do it wrong. Who knows.
amistre64
  • amistre64
when I decomped it I allowed the square root to be a negative number... so I had to go another route...
anonymous
  • anonymous
OH wow.
amistre64
  • amistre64
if I did it right...gonna have to try to derive it down; \[\frac{\tan^{-1}(x^2)}{2}\]
amistre64
  • amistre64
I think thats it :)\[Dx (\frac{\tan^{-1}(x^2)}{2}) \rightarrow \frac{2x}{2[(x^2)^2+1]}\]
amistre64
  • amistre64
which = x/(x^4 +1) :)
amistre64
  • amistre64
\[\frac{\tan^{-1}(x^2)}{2} + C\]
anonymous
  • anonymous
Okay...I see I see...Thank you so much for all of your effort and help. :)
amistre64
  • amistre64
youre welcome :)

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