Question

求积分：x的4次方加1分之1

Answer 1

这个用因式分解法即可,其中a=根号2,b=根号3,方便书写
∫dx/(1+x^4)
=∫dx/[(x^2+ax+1)(x^2-ax+1)]
=∫dx[(1/2a)x+(1/2)]/(x^2+ax+1)+∫dx[(-1/2a)x+(1/2)]/(x^2-ax+1)
=∫(x+a)dx/[2a(x^2+ax+1)]-∫(x-a)dx/[2a(x^2-ax+1)]
=∫(x+a/2+a/2)dx/[2a(x^2+ax+1)]-∫(x-a/2-a/2)dx/[2a(x^2-ax+1)]
=∫(x+a/2)dx/[2a(x^2+ax+1)]-∫(x-a/2)dx/[2a(x^2-ax+1)]+∫dx/[4(x^2+ax+1)]-∫adx/[4(x^2-ax+1)]
=(1/2a)ln(x^2+ax+1)-(1/2a)ln(x^2-ax+1)+(1/2b)arctan(2x/b+a/b)-(1/2b)arctan(2x/b-a/b)+C