Question

问怎么求 ∫x的平方/（1+x的4次方）dx

Answer 1

∫x^2dx/(1+x^4)
=∫x^2dx/[(1+x^2)^2-2x^2]
=∫x^2dx/[(1+x^2-√2x)(1+x^2+√2x)]
=(1/(2√2))∫2√2xdx^2/[(1+x^2-√2x)(1+x^2+√2x)]
=(1/(2√2))[∫dx^2/(1+x^2-√2x)- ∫dx^2/(1+x^2+√2x)]
=(1/√2)[∫xdx/√(1+x^2-√2x) -∫xdx/(1+x^2+√2x)]
∫xdx/(1+x^2-√2x)=∫(x-√2/2)dx/(1+x^2-√2x)+ (√2/2)∫dx/(1+x^2-√2x)
=(1/2)ln|1+x^2-√2x| +(√2/2)∫dx/[(x-√2/2)^2+3/4]
=(1/2)ln|1+x^2-√2x|+√(2/3)∫d(2x-√(2/3))/[(2x-√(2/3))^2+1]
=(1/2)ln|1+x^2-√2x|+√(2/3)arctan(2x-√(2/3))
∫xdx/(1+x^2+√2x)=(1/2)ln|1+x^2+√2x| -(√(2/3)arctan(2x+√(2/3))

∫√x^2dx/(1+x^4)
=(1/2√2)ln[|1+x^2-√2x|/|1+x^2+√2x|] +(1/√3)arctan(2x-√(2/3))-(1/√3)artcan(2x+√(2/3))+C