## anonymous 5 years ago 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?

1. amistre64

you wanna indefinite integral both of them right?

2. anonymous

Yes.

3. amistre64

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

4. amistre64

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

5. anonymous

Right...

6. amistre64

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

7. amistre64

by second I mean first :)

8. anonymous

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

9. anonymous

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

10. anonymous

$\int\limits 1/u$

11. amistre64

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

12. amistre64

..du

13. amistre64

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

14. amistre64

sounds good, looks bad; decompose the fraction :)

15. amistre64

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

16. amistre64

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

17. amistre64

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

18. anonymous

Yeah, I was about to say that you were correct. So, would you mind walking me through how to solve the second one?

19. amistre64

and by second do you mean first?

20. anonymous

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

21. 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

22. amistre64

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

23. anonymous

yes.

24. amistre64

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

25. anonymous

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

26. 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.

27. amistre64

was the decomposition one right? or do we know?

28. anonymous

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

29. amistre64

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

30. amistre64

running into problems with it ...

31. amistre64

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

32. anonymous

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

33. amistre64

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

34. anonymous

OH wow.

35. amistre64

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

36. amistre64

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

37. amistre64

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

38. amistre64

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

39. anonymous

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

40. amistre64

youre welcome :)