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integrate 1/(x^4 +1)
Dear Gurvinder,
1 / ( x4+1) = (x2+1) - (x2-1) ] /2(x4+1) dx
(x2+1)/2(x4+1) dx = (1+ 1/x2)/2[(x-1/x)2 + 2] dx ( divide by x2 and add 2 and -2 in denominator to make the whole square)
now pur x-1/x = y or (1+ 1/x2)dx = dy
put above value in integral and solve
similarly, solve intergral : -(x2-1)/2(x4+1) dx
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Askiitians Expert
Anil Pannikar
IIT Bombay
∫dx/x4+1=1/2∫x2+1+1-x2./1+x4
=1/2∫x2+1/x2+1.dx+1/2∫1-x2/x4+1dx
=1/2∫(1+1/x2)/(x2+1/x2) + 1/2∫(1/x2-1)/1/x2+x2)
putting x-1/x =z and 1+ 1/x2=dz for first term ie [1/2∫(1+1/x2)/(x2+1/x2)] and in second termx+1/x as t and 1-1/x2 as dt
then on substituting and solving we get 1/23/2tan-11/21/2(x-1/x)-1/25/2log[(x2-2x1/2+1)/(x2+2x1/2+1)] got it all the best
sorry no idea :(
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