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 1/2^3/2tan^-11/2^1/2(x-1/x)-1/2^5/2log[(x²-2x^1/2+1)/(x²+2x^1/2+1)] got it all the best

sorry no idea :(