令x² = tanθ,2x dx = sec²θ dθ
∫ x/(1 + x^4) dx
= ∫ x/(1 + tan²θ) • sec²θ dθ/(2x)
= (1/2)∫ sec²θ/sec²θ dθ
= (1/2)θ + C
= (1/2)arctan(x²) + C
求x/1+x^4的不定积分用第二换元法
令x² = tanθ,2x dx = sec²θ dθ
∫ x/(1 + x^4) dx
= ∫ x/(1 + tan²θ) • sec²θ dθ/(2x)
= (1/2)∫ sec²θ/sec²θ dθ
= (1/2)θ + C
= (1/2)arctan(x²) + C