Question

求解答一道积分题~

Answer 1

1/(x^4+1)=(√2/4x+1/2)/(x^2+√2x+1)+(-√2/4x+1/2)/(x^2-√2x+1)

∫[(√2/4)x+1/2]/(x^2+√2x+1)]dx
=(√2/8)[∫(2x+√2)/(x^2+√2x+1)dx+∫√2/(x^2+√2x+1)dx
=(√2/8)[ln(x^2+√2x+1)+2arctan(√2x+1)]+C
后半个式子方法相同
所以∫1/(x^4+1)dx
=(√2/8)[ln(x^2+√2x+1)-ln(x^2-√2x+1)+2arctan(√2x+1)+2arctan(√2x-1)]+C

Answer 2

a=x-1/x
a2=x2-2+1/x2
x2+1/x2=a+2
则a+2+2a-1=0
a2+2a+1=0
(a+1)2=0
a=-1

x-1/x=-1
两边乘x
x2+x-1=0
x=(-1±√5)/2