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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?
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?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you wanna indefinite integral both of them right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the last one just split the fraction: x^4/x +1/x and integrate

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} \frac{x^4}{x} + \frac{1}{x} dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the second one we can try subsituting u=x^4+1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0by second I mean first :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I gotcha :) I figured as much...first and second...same thing. right? haha!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so, on the first one you end up with int(1/u)du?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{} \frac{1}{4u(u1)^{1/2}}\]this is what I get :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0put the u under the radial and take out the 1/4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0sounds good, looks bad; decompose the fraction :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{4}\int\limits_{}\frac{1}{u}du  \int\limits_{} \frac{1}{(u+1)^{1/2}}du\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0umm... what did I do lol...hold up on that

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, thats right, I couldnt recall where I got (u1)^1/2 from

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, I was about to say that you were correct. So, would you mind walking me through how to solve the second one?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and by second do you mean first?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha! yes, the first one we talked about anyway.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0ok.... 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
 5 years ago
Best ResponseYou've already chosen the best response.0like denominators simply merge and you work the tops; well, they can be split up just as easily

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we get: x^3 + 1/x which integrates to: x^4/4 + ln(x) + C

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh! I understand! Thank you so much. Now, do we need to add a plus c to the second one we did?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0whenever 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
 5 years ago
Best ResponseYou've already chosen the best response.0was the decomposition one right? or do we know?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I don't think we know, unless we can infer that it is from the original problem I gave.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0let me redo that one on paper to see if I got it right....

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0running into problems with it ...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i might just be over thinking it, let me try trig substitution...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I didn't really notice anything wrong, but I could just be trying to do it wrong. Who knows.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0when I decomped it I allowed the square root to be a negative number... so I had to go another route...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if I did it right...gonna have to try to derive it down; \[\frac{\tan^{1}(x^2)}{2}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0I think thats it :)\[Dx (\frac{\tan^{1}(x^2)}{2}) \rightarrow \frac{2x}{2[(x^2)^2+1]}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0which = x/(x^4 +1) :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\tan^{1}(x^2)}{2} + C\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay...I see I see...Thank you so much for all of your effort and help. :)
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