1cn write it as: 1/2 ∫ [( x2+1) - (x2-1)] / (1+x4) dx
now, separate the integrals and then,divide x2 in both numerator and denominator in both the integrals.
we'll get : = 1/2 [ ∫( 1+ 1/x2) dx /(x- 1/x)2+2 ) - ∫(1- 1/x2)dx / ((x+ 1/x)2-2)]
now,substitute (x- 1/x)=t,in the first one, and (x+ 1/x)=t in the 2nd one.
expression becomes 1/2[ ∫dt /t2+2 - ∫dt /t2-2 ]
now, its integrable, i think.
the final answer would be = 1/2 [ (1/√2) tan-1((x-1/x) / √2 ) - {1/2√2 ln| (x+1/x - √2) / (x+1/x) + √2|} ]
