原式=∫x³dx/[x^4√(2x^8-2x^4+1)]
=1/4∫d(x^4)/[x^4√(2(x^4-1/2)²+1/2)]
=1/(4√2)∫d(x^4)/[x^4√((x^4-1/2)²+(1/2)²)]
=1/(4√2)∫(1/2)sec²θdθ/[((1+tanθ)/2)*(secθ/2)]
(设x^4-1/2=(1/2)tanθ,则x^4=(1+tanθ)/2,d(x^4)=(1/2)sec²θdθ)
=1/(2√2)∫secθdθ/(1+tanθ)
=1/(2√2)∫dθ/(sinθ+cosθ)
=1/(2√2)∫[2dt/(1+t²)]/[2t/(1+t²)+(1-t²)/(1+t²)]
(设t=tan(θ/2),则sinθ=2t/(1+t²),cosθ=(1-t²)/(1+t²),dθ=2dt/(1+t²))
=1/√2∫[dt/(1+2t-t²)
=1/4∫[1/(t-1+√2)-1/(t-1-√2)]dt
=1/4ln│(t-1+√2)/(t-1-√2)│+C (C是积分常数)
=1/4ln│(tan(θ/2)-1+√2)/(tan(θ/2)-1-√2)│+C
（其中x^4-1/2=(1/2)tanθ）.
说明：此题最后结果不要还原回去,因为还原回去太复杂了.
