Question

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?

Answer 1

you wanna indefinite integral both of them right?

Answer 2

Yes.

Answer 3

the last one just split the fraction:

x^4/x +1/x and integrate

Answer 4

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

Answer 5

Right...

Answer 6

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

Answer 7

by second I mean first :)

Answer 8

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

Answer 9

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

Answer 10

\[\int\limits 1/u\]

Answer 11

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

Answer 12

..du

Answer 13

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

Answer 14

sounds good, looks bad; decompose the fraction :)

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

and by second do you mean first?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

yes.

Answer 24

we get:

x^3 + 1/x which integrates to:

x^4/4 + ln(x) + C

Answer 25

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

Answer 26

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.

Answer 27

was the decomposition one right? or do we know?

Answer 28

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

Answer 29

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

Answer 30

running into problems with it ...

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

OH wow.

Answer 35

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

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

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

Answer 40

youre welcome :)