一道高数题,∫[1/(1+x^4)]dx

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一道高数题,∫[1/(1+x^4)]dx一道高数题,∫[1/(1+x^4)]dx一道高数题,∫[1/(1+x^4)]dx∫[1/(1+x^4)]dx=1/2∫[(x^2+1)-(x^2-1)]/(1+

一道高数题,∫[1/(1+x^4)]dx
一道高数题,
∫[1/(1+x^4)]dx

一道高数题,∫[1/(1+x^4)]dx
∫[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)/(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/{[(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