A problem on integration
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Yes, this is a homework question. It is really supposed to be a review of the latter part of calc I where integration is introduced. I've been able to go through all the problems seemlessly so far except integrating this one:
∫(t^3/(t^6+1))dt
The problem seems simple enough, and in keeping with the rules I'm not looking for answers, just a nudge in the right direction please. I tried substitution and dividing by various powers of t for the numerator and denominator and its a no go. An appropriate answer would be telling me if there is a technique I should google on please or perhaps one critical step I'm not implementing. Thank you.
[edited by - nervo on January 23, 2004 10:49:10 PM]
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quote:Original post by SiCrane
Ok, a hint: substituting x=t^3 should give you something familiar looking.
I just fixed the problem I had above. I've already tried what you said though.... if say u = t^3 then du = 3t^2dx
So...I don't see how I can play with the variables to get anything useful. One thing I was trying to do was get a single term on the bottom so I could divide through how many terms would be on the top...But again, no go.
wait a sec....unless you're saying subsituting x = t^3 would give me x/(x^2+1)...which would be something close looking to an inverse tangent. Hmm..
[edited by - nervo on January 23, 2004 10:53:56 PM]
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Perhaps I took your hint wrong. I ended doing something like a double subsitution which I haven't done nor do I know if it's a technique that is used
It seemed too good to be true..If I used x = t^3 and had:
x/(x^2+1) then using subsitution for that with u = x^2 + 1 and du/2 = xdx and having 1/2∫du/u and then integrating that into 1/2ln|x^2+1| and finally subsituting t^3 into x to get 1/2ln|t^6+1|
Well...I'm dead wrong, because differentiating the final result does not take me back to the original function
I'm toying around with it so much that I must be using some real funky methods just to get an answer at this point.
[edited by - nervo on January 23, 2004 11:28:08 PM]
It seemed too good to be true..If I used x = t^3 and had:
x/(x^2+1) then using subsitution for that with u = x^2 + 1 and du/2 = xdx and having 1/2∫du/u and then integrating that into 1/2ln|x^2+1| and finally subsituting t^3 into x to get 1/2ln|t^6+1|
Well...I'm dead wrong, because differentiating the final result does not take me back to the original function
I'm toying around with it so much that I must be using some real funky methods just to get an answer at this point. [edited by - nervo on January 23, 2004 11:28:08 PM]
According to Maple, the answer has three large complicated terms, two of which are ln(stuff)''s and the other is an arctan(stuff) where stuff are polynomials of order 2 or 4. Is this a challenge problem? Perhaps it''s an error... It doesn''t seem like something you would be expected to be able to solve by hand on a test / assignment...
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quote:Original post by Geoff the Medio
According to Maple, the answer has three large complicated terms, two of which are ln(stuff)''s and the other is an arctan(stuff) where stuff are polynomials of order 2 or 4. Is this a challenge problem? Perhaps it''s an error... It doesn''t seem like something you would be expected to be able to solve by hand on a test / assignment...
It was a homework question. Maybe the teacher made an slight error and in doing so made it harder. I''ll need to verify that. Unless she''s sadistic. Integral calculus is like nothing I''ve done before in that there are no step by step rules that can conveniently apply to all problems. It seems like alot of experience is needed, and I''m doing all the problems in the book plus Shaums calculus.
Ok, I think I was able to it. It ain''t pretty. My answer differs significantly from the Integrators, but I think either I have made simple mistake, or I am missing some clever re-arrangement that makes them the same. So I am fairly sure the following method is along the right lines:
I used the substitution x=t^2. From there its partial fractions, and then a reach to the formula booklet for the last part. A nice integral, it covers practically everything I have learnt on integration. It does involve lns and arctans.
I used the substitution x=t^2. From there its partial fractions, and then a reach to the formula booklet for the last part. A nice integral, it covers practically everything I have learnt on integration. It does involve lns and arctans.
OK, I''ve got it now. THe above method is correct. You do also get high power things, but you can factorize them and seperate out the factors as seperate lns.
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