Question

高数换元积分法求积分f xdx\１+x平方+x四次方.因为看了解析也没有看懂.

Answer 1

=(1/2)∫dx^2/(1+x^2+x^4) (t=x^2)

=(1/2)∫dt/(t^2+t+1)

=(1/2)∫dt/[(t+1/2)^2+3/4)

=(1/2)(2/√3)arctan[(t+1/2)/(√3/2)]+C

=(1/√3)arctan[(2t+1)/√3]+C

=(1/√3)arctan[(2x^2+1)/√3]+C