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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

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

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

and by second do you mean first?

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

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?

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.

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 :)