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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: Tough Integral question
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vivek_aero (0)

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Tough Integral question
    
vivek_aero (0)

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Re:Tough Integral question
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vivek_aero (0)

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How to integrate (x^4+1)/(x^6+1) i tried substituting tant but ended up getting tan^2/3 which is tough to evaluate .is there an elegant way of evaluating this.i also tried partial fractions but that is horrible.
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kushi12 (454)

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int (X^2)^2+1/(X^2)^3+1


APLY FORMULA


A^2+B^2((A+B)^2-2AB)IN NUMURATOR


AND A^3+B^3 IN DENOMRATOR(A+B)(A^2+B^2-AB)


NOW U GET INT X^2+1/X^4-X^2+1 - INT 2 X^2/(X^2+1)(X^4+1-X^2)


PUT X^2=T


INT T+1/T^2-T+1 -INT2T/(T+1)(T^2-T+1)


NOW U CAN APPLY PARTIAL IF CAN'T APPLY TELL ME I DO


 

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vnkt.swaroop (419)

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 integrate this question and tell whether it is integrable or not?

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determinedforiit (552)

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 it is integrable but a very long answer. see wolfram integrator.


Only change is eternal
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rudra.panda (2733)

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cannot be integrated without a calculator(I think so)


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