求不定积分∫dx/(1+x^4)

---

∫[1/（1+x^4）]dx

= 1/2∫[(x^2+1)-(x^2-1)]/（1+x^4）dx

= 1/2 {∫(x^2+1)/(1+x^4) dx - ∫(x^2-1)/兆激裤（1+x^4）dx }

= 1/2 {∫(1+1/x^2)dx /（x^2+1/x^2） - ∫(1-1/x^2)dx/（x^2+1/x^2）}

= 1/2 {∫d(x-1/x) /[(x-1/x)^2+2] - ∫d(x+1/x) /[(x+1/x)^2 -2] }

= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1} - ∫d(x+1/x) /[(x+1/x)^2 -2] }

= 1/2 { 1/√2 ∫d[(x-1/x) /√2] /{[(x-1/x)/√2]^2+1}
- 1/2√2 ∫d[(x+1/x) /√2] [ 1/{[(x+1/x)/√2] -1} - 1/铅码{[(x+1/x)/√2] +1 }]

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

【或者，使用待定系族简数法,但较繁琐：】

∫[1/（1+x^4）]dx

=∫ 1/[(x^2-x√2+1)*(x^2+x√2 +1)]dx

=∫ { [ax+b]/[(x^2-x√2+1) + [cx+d]/(x^2+x√2 +1)] }dx

---

点祥凳橡粗李击放大、再点击谨旁再放大：

---