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x-2-3-x-6-x-2-1-dx-Is-there-any-special-method-of-decomposition-other-than-the-use-of-partial-fractions-




Question Number 105411 by Ar Brandon last updated on 28/Jul/20
∫((x^2 +3)/(x^6 (x^2 +1)))dx  Is there any special method of decomposition  other than the use of partial fractions ?
x2+3x6(x2+1)dxIsthereanyspecialmethodofdecompositionotherthantheuseofpartialfractions?
Answered by Dwaipayan Shikari last updated on 28/Jul/20
∫((x^2 +1)/(x^6 (x^2 +1)))+(2/(x^6 (x^2 +1)))dx  ∫(1/x^6 )+2∫(1/x^4 )((1/x^2 )−(1/(x^2 +1)))dx  3∫(1/x^6 )−2∫(1/(x^4 (x^2 +1)))dx  3∫(1/x^6 )−2∫(1/x^2 )((1/x^2 )−(1/(x^2 +1)))dx  3∫(1/x^6 )−2∫(1/x^4 )+2∫(1/(x^2 (x^2 +1)))dx  3∫(1/x^6 )−2∫(1/x^4 )+2∫(1/x^2 )−2∫(1/(x^2 +1))dx  −(3/(5x^5 ))+(2/(3x^3 ))−(2/x)−2tan^(−1) x+C
x2+1x6(x2+1)+2x6(x2+1)dx1x6+21x4(1x21x2+1)dx31x621x4(x2+1)dx31x621x2(1x21x2+1)dx31x621x4+21x2(x2+1)dx31x621x4+21x221x2+1dx35x5+23x32x2tan1x+C
Commented by Dwaipayan Shikari last updated on 28/Jul/20
I think so . Is it a better way that you have mentioned?
Ithinkso.Isitabetterwaythatyouhavementioned?
Commented by Dwaipayan Shikari last updated on 28/Jul/20
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Commented by Ar Brandon last updated on 28/Jul/20
Perfect, bro. Exactly what I needed. Thanks��
Commented by Ar Brandon last updated on 28/Jul/20
(1/(x^2 (x^2 +1)))=((x^2 +1−x^2 )/(x^2 (x^2 +1)))=((x^2 +1)/(x^2 (x^2 +1)))−(x^2 /(x^2 (x^2 +1)))                       =(1/x^2 )−(1/(x^2 +1)) ,
1x2(x2+1)=x2+1x2x2(x2+1)=x2+1x2(x2+1)x2x2(x2+1)=1x21x2+1,

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