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?

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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
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you wanna indefinite integral both of them right?
Yes.
the last one just split the fraction: x^4/x +1/x and integrate

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\[\int\limits_{} \frac{x^4}{x} + \frac{1}{x} dx\]
Right...
the second one we can try subsituting u=x^4+1
by second I mean first :)
I gotcha :) I figured as much...first and second...same thing. right? haha!
so, on the first one you end up with int(1/u)du?
\[\int\limits 1/u\]
\[\int\limits_{} \frac{1}{4u(u-1)^{1/2}}\]this is what I get :)
..du
put the u under the radial and take out the 1/4
sounds good, looks bad; decompose the fraction :)
\[\frac{1}{4}\int\limits_{}\frac{1}{u}du - \int\limits_{} \frac{1}{(u+1)^{1/2}}du\]
umm... what did I do lol...hold up on that
yeah, thats right, I couldnt recall where I got (u-1)^1/2 from
Yeah, I was about to say that you were correct. So, would you mind walking me through how to solve the second one?
and by second do you mean first?
Haha! yes, the first one we talked about anyway.
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
like denominators simply merge and you work the tops; well, they can be split up just as easily
yes.
we get: x^3 + 1/x which integrates to: x^4/4 + ln(x) + C
Oh! I understand! Thank you so much. Now, do we need to add a plus c to the second one we did?
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.
was the decomposition one right? or do we know?
I don't think we know, unless we can infer that it is from the original problem I gave.
let me redo that one on paper to see if I got it right....
running into problems with it ...
i might just be over thinking it, let me try trig substitution...
I didn't really notice anything wrong, but I could just be trying to do it wrong. Who knows.
when I decomped it I allowed the square root to be a negative number... so I had to go another route...
OH wow.
if I did it right...gonna have to try to derive it down; \[\frac{\tan^{-1}(x^2)}{2}\]
I think thats it :)\[Dx (\frac{\tan^{-1}(x^2)}{2}) \rightarrow \frac{2x}{2[(x^2)^2+1]}\]
which = x/(x^4 +1) :)
\[\frac{\tan^{-1}(x^2)}{2} + C\]
Okay...I see I see...Thank you so much for all of your effort and help. :)
youre welcome :)

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