求函数f（x）=1/(1+x4)在零到正无穷上的定积分

∵∫dx/(1+x^4)=∫dx/[(x²+√2x+1)(x²-√2x+1)]
=(1/(4√2))∫[(2x+√2+√2)/(x²+√2x+1)-(x-√2-√2)/(x²-√2x+1)]dx
=(1/(4√2))[∫(2x+√2)dx/(x²+√2x+1)-∫(x-√2)dx/(x²-√2x+1)
+√2∫dx/(x²+√2x+1)+√2∫dx/(x²-√2x+1)]
=(1/(4√2))[∫d(x²+√2x+1)/(x²+√2x+1)-∫d(x²-√2x+1)/(x²-√2x+1)
+√2∫dx/((x+1/√2)²+1/2)+√2∫dx/((x-1/√2)²+1/2)]
=(1/(4√2))[ln(x²+√2x+1)-ln(x²-√2x+1)+2arctan(√2x+1)+2arctan(√2x-1)]│
=(1/(4√2))[ln((x²+√2x+1)/(x²-√2x+1))+2arctan(√2x+1)+2arctan(√2x-1)]│
=(1/(4√2))[ln((t²+√2t+1)/(t²-√2t+1))+2arctan(√2t+1)+2arctan(√2t-1)
-2arctan(1)-2arctan(-1)]
=(1/(4√2))[ln((t²+√2t+1)/(t²-√2t+1))+2arctan(√2t+1)+2arctan(√2t-1)-π/2+π/2]
=(1/(4√2))[ln((t²+√2t+1)/(t²-√2t+1))+2arctan(√2t+1)+2arctan(√2t-1)]
∴原式=lim(t->+∞)∫dx/(1+x^4)
=lim(t->+∞){(1/(4√2))[ln((t²+√2t+1)/(t²-√2t+1))+2arctan(√2t+1)+2arctan(√2t-1)]}
=lim(t->+∞){(1/(4√2))[ln((1+√2/t+1/t²)/(1-√2/t+1/t²))+2arctan(√2t+1)+2arctan(√2t-1)]}
=(1/(4√2))[ln((1+0+0)/(1-0+0))+2(π/2)+2(π/2)]
=(1/(4√2))(2π)
=π/(2√2).

这属于一个反常积分问题，并且属于无穷限问题；
当x趋近于无穷大时，f(x)趋近于0
而x=0时，f(x)=1;
所以结果为：-1