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?

- anonymous

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- amistre64

you wanna indefinite integral both of them right?

- anonymous

Yes.

- amistre64

the last one just split the fraction:
x^4/x +1/x and integrate

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## More answers

- amistre64

\[\int\limits_{} \frac{x^4}{x} + \frac{1}{x} dx\]

- anonymous

Right...

- amistre64

the second one we can try subsituting u=x^4+1

- amistre64

by second I mean first :)

- anonymous

I gotcha :) I figured as much...first and second...same thing. right? haha!

- anonymous

so, on the first one you end up with int(1/u)du?

- anonymous

\[\int\limits 1/u\]

- amistre64

\[\int\limits_{} \frac{1}{4u(u-1)^{1/2}}\]this is what I get :)

- amistre64

..du

- amistre64

put the u under the radial and take out the 1/4

- amistre64

sounds good, looks bad; decompose the fraction :)

- amistre64

\[\frac{1}{4}\int\limits_{}\frac{1}{u}du - \int\limits_{} \frac{1}{(u+1)^{1/2}}du\]

- amistre64

umm... what did I do lol...hold up on that

- amistre64

yeah, thats right, I couldnt recall where I got (u-1)^1/2 from

- 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

and by second do you mean first?

- anonymous

Haha! yes, the first one we talked about anyway.

- 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

like denominators simply merge and you work the tops; well, they can be split up just as easily

- anonymous

yes.

- amistre64

we get:
x^3 + 1/x which integrates to:
x^4/4 + ln(x) + C

- anonymous

Oh! I understand! Thank you so much. Now, do we need to add a plus c to the second one we did?

- 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

was the decomposition one right? or do we know?

- anonymous

I don't think we know, unless we can infer that it is from the original problem I gave.

- amistre64

let me redo that one on paper to see if I got it right....

- amistre64

running into problems with it ...

- amistre64

i might just be over thinking it, let me try trig substitution...

- anonymous

I didn't really notice anything wrong, but I could just be trying to do it wrong. Who knows.

- amistre64

when I decomped it I allowed the square root to be a negative number... so I had to go another route...

- anonymous

OH wow.

- amistre64

if I did it right...gonna have to try to derive it down; \[\frac{\tan^{-1}(x^2)}{2}\]

- amistre64

I think thats it :)\[Dx (\frac{\tan^{-1}(x^2)}{2}) \rightarrow \frac{2x}{2[(x^2)^2+1]}\]

- amistre64

which = x/(x^4 +1) :)

- amistre64

\[\frac{\tan^{-1}(x^2)}{2} + C\]

- anonymous

Okay...I see I see...Thank you so much for all of your effort and help. :)

- amistre64

youre welcome :)

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