答：

　　我曾经答过一样的题.

　　原式

　　=∫(x^2+1)/[2(x^4+1)]dx-∫(x^2-1)/[2(x^4+1)]dx

　　=1/2∫(1+1/x^2)/(x^2+1/x^2)dx-1/2∫(1-1/x^2)/(x^2+1/x^2)dx

　　=1/2∫d(x-1/x)/[(x-1/x)^2+2]-1/2∫d(x+1/x)/[(x+1/x)^2-2]

　　=1/4∫d(x-1/x)/[(x-1/x)^2/2+1]-1/2∫d(x+1/x)/[(x+1/x+√2)(x+1/x-√2)]

　　=√2/4*arctan[(x-1/x)/√2]-1/4∫d(x+1/x)/(x+1/x+√2)-1/4∫d(x+1/x)/(x+1/x-√2)

　　=√2/4*arctan[(x-1/x)/√2]-1/4*ln|x+1/x+√2|-1/4*ln|x+1/x-√2|+C