1/{x*(x^4+2x^2-1)}的不定积分

Question

1/{x*(x^4+2x^2-1)}的不定积分

Answer 1

∫dx/[x*(x^4+2x^2-1)]=∫dx/[x*(x^2+1)^2-2]
=∫dx/[x*(x^2+1-√2)(x^2+1+√2)]
=(1/2√2)∫dx/[x*(x^2+1-√2) -(1/2√2)∫dx/[x*(x^2+1+√2)]
∫dx/[x*(x^2+1-√2)=1/(1-√2)∫[(x^2+1-√2)-x^2]dx/[x*(x^2+1-√2)]
=1/(1-√2)[∫dx/x-(1/2)∫d(x^2+1-√2)/(x^2+1-√2)
=1/(1-√2)[ln|x| -(1/2)ln|x^2+1-√2|]
∫dx/[x*(x^2+1+√2)=1/(1+√2)∫[(x^2+1+√2)-x^2]dx/[x*(x^2+1+√2)]
=(1/(1+√2)ln|x|-(1/2)ln|x^2+1+√2|
∫dx/[x*(x^4+2x^2-1)]=1/(2√2-4) *ln|x| - 1/(4√2-8)ln|x^2+1-√2|
-1/(2√2+4) *ln|x| + 1/(4√2+8)ln|x^2+1+√2| +C